Coordinate Geometry

Start with a question that has been set, in one form or another, for decades:

Find the shortest distance between the line $y-x=1$ and the curve $x=y^2$.

The obvious attack is to take a general point $(x,y)$ on the parabola, write the distance to the line, and minimise. That is a two-variable problem with a square root in it, and it goes nowhere pleasant.

Now watch what one choice does. Every point on $x=y^2$ can be written $(t^2,t)$ — one parameter, not two. The line is $x-y+1=0$, so the distance from that point is

$D=\frac{|t^2-t+1|}{\sqrt{2}}$

The quadratic $t^2-t+1$ has discriminant $1-4=-3<0$, so it never touches zero and never changes sign. The modulus can be dropped. What is left is a quadratic in one variable, minimised at $t=\frac{1}{2}$, where it equals $\frac{3}{4}$. So $D=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$.

Four lines. Nothing clever happened — the curve was written in the form that had one unknown instead of two, and a hard minimisation turned into a school-level one.

That is the whole chapter. Coordinate geometry is a translation service between geometry and algebra, and the exam is not testing whether you can translate. It is testing whether you pick the translation that leaves the least algebra behind. Everything below is either a translation worth knowing or a rule for choosing between them.

Part I — Points and lines

1. Your one free choice

Before any calculation, you get to place the origin and point the axes wherever you like. You get that choice once. Spend it making coordinates zero.

If a triangle has a right angle, put the right angle at the origin and the two legs along the axes: the three vertices become $(0,0)$, $(p,0)$, $(0,q)$ and two of the six coordinates die. If a figure has an axis of symmetry, put it on the $x$-axis so points pair up as $(\alpha ,\pm \beta )$. If a chord matters more than the curve, aim an axis along the chord.

One consequence worth carrying: in a right-angled triangle the orthocentre is the right-angle vertex itself. The altitude from the right angle is one line; the other two altitudes are the legs themselves, since each leg is already perpendicular to the other. All three meet where the legs meet. No calculation, and it has been the whole content of more than one exam question.

2. The three results everything else is built from

Given $A(x_1,y_1)$ and $B(x_2,y_2)$, the distance is

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

which is Pythagoras on the horizontal and vertical gaps, nothing more.

For the section formula, let $P$ divide $AB$ internally in the ratio $m:n$. Drop perpendiculars from $A$, $P$ and $B$ onto the $x$-axis. The two right triangles formed are similar, so horizontal gaps are in the same ratio as the segments: $\frac{x-x_1}{x_2-x}=\frac{m}{n}$. Cross-multiplying, $n(x-x_1)=m(x_2-x)$, so $nx+mx=m{x}_2+n{x}_1$. The same argument on the $y$-axis gives the other coordinate:

$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

Put $m=n$ and you have the midpoint. Put $n$ negative and you have external division — which is the only thing that changes for the external case, so there is no second formula to learn.

The centroid is the point dividing each median in the ratio $2:1$ from the vertex, and running the section formula on that gives $G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$. Note what this means for area questions: joining the centroid to the three vertices splits the triangle into three pieces of equal area.

So if a question says "$P$ lies inside triangle $ABC$ and the three triangles $PAB$, $PBC$, $PCA$ have equal areas", it has told you $P$ is the centroid. Read that as a gift.

3. Distance from a point to a line, and why the formula looks like that

Most candidates memorise this one and never see where it comes from, which is why they cannot reconstruct it when the line is given in another form.

Take the line $ax+by+c=0$. If $(x,y)$ and $(x^{\prime },y^{\prime })$ both lie on it, subtracting the two equations gives $a(x-x^{\prime })+b(y-y^{\prime })=0$. That says the vector $(a,b)$ is perpendicular to every direction along the line. So the coefficients are a normal vector — that single fact is what the formula is made of.

Now let $Q(x_0,y_0)$ be any point on the line and $P(x_1,y_1)$ the point you are measuring from. The distance is the length of the shadow of $\overrightarrow{QP}$ on the unit normal:

$d=\left|\overrightarrow{QP}\cdot \hat{n}\right|=\frac{|a(x_1-x_0)+b(y_1-y_0)|}{\sqrt{a^2+b^2}}$

Since $Q$ is on the line, $a{x}_0+b{y}_0=-c$, and substituting that in collapses everything to

$\text{distance}=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

Two things follow immediately, and both are worth more than the formula itself.

The distance between the parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$ — take any point on one and apply the result.

And the sign of $a{x}_1+b{y}_1+c$, before the modulus, tells you which side of the line the point is on. Two points share a side exactly when the two expressions share a sign. That test is faster than any diagram, and it is how you check whether a point lies inside a triangle.

Illustration 1

The perpendicular distance from the origin to the normal to the curve $x^2+2xy-3y^2=0$ at the point $(2,2)$. (JEE previous-year question)

The "curve" is a red herring — a homogeneous quadratic in $x$ and $y$ always factors into two straight lines through the origin. Here $x^2+2xy-3y^2=(x+3y)(x-y)$, so the equation is the pair of lines $x+3y=0$ and $x-y=0$.

The point $(2,2)$ satisfies the second, so the relevant line is $y=x$, whose normal at $(2,2)$ has slope $-1$: $x+y=4$. The distance from the origin is $\frac{|0+0-4|}{\sqrt{1+1}}=\frac{4}{\sqrt{2}}$. Answer: $2\sqrt{2}$.

4. Angle between two lines — and the version that never breaks

If two lines make angles ${\theta }_1,{\theta }_2$ with the $x$-axis then $m_1=\tan {\theta }_1$, $m_2=\tan {\theta }_2$, and the angle between them is ${\theta }_1-{\theta }_2$. Expanding $\tan ({\theta }_1-{\theta }_2)$ gives

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Parallel means the numerator vanishes, $m_1=m_2$. Perpendicular means the denominator vanishes, $m_1{m}_2=-1$.

That last statement hides a trap. A vertical line has no slope at all, so "$m_1{m}_2=-1$" cannot detect that $x=3$ and $y=5$ are perpendicular. The normal-vector version has no such gap: $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are perpendicular exactly when $a_1{a}_2+b_1{b}_2=0$, and parallel exactly when $a_1{b}_2-a_2{b}_1=0$. Use this version whenever the lines arrive in general form, which in an exam is most of the time.

For choosing a form to write a line in: use $y-y_1=m(x-x_1)$ when you have a point and a direction, $\frac{x}{p}+\frac{y}{q}=1$ when the question mentions intercepts (it hands you the intercepts directly), and $x\cos \alpha +y\sin \alpha =p$ when the question is about the perpendicular from the origin. Choosing the form that matches the given data is worth a minute per question.

5. The family of lines, or how to avoid finding the intersection

Suppose $L_1=0$ and $L_2=0$ meet at a point $P$. Then for any $\lambda$, the equation

$L_1+\lambda L_2=0$

is satisfied at $P$, because both terms are separately zero there. It is also linear, so it is a line. Therefore $L_1+\lambda L_2=0$ describes every line through $P$ (bar $L_2$ itself), with $\lambda$ selecting which one.

The payoff: when a question asks for the line through the intersection of two given lines that also satisfies one more condition, you never have to compute the intersection. Write $L_1+\lambda L_2=0$, impose the extra condition, solve for $\lambda$. One unknown instead of a simultaneous pair, and no fractions to carry.

The same idea works one level up. If $S_1=0$ and $S_2=0$ are circles, $S_1-S_2=0$ is linear — the squared terms cancel — and it is the line through their two intersection points, the radical axis. You get the common chord without ever solving for either intersection point.

6. Concurrence, and area without a diagram

Three lines $a_{i}x+b_{i}y+c_i=0$ are concurrent when the determinant of their coefficients vanishes:

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

The reason is that concurrence means the three equations have a common solution $(x,y)$, so the system $a_{i}x+b_{i}y+c_i\cdot 1=0$ has a non-trivial solution $(x,y,1)$ — and a homogeneous $3\times 3$ system has a non-trivial solution exactly when its determinant is zero.

The area of the triangle on $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

$\text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

Set it to zero and you have the collinearity test. Set it to a given value and you get a pair of parallel lines as the locus of the third vertex — one for each sign before the modulus is applied. Exam questions exploit that: "the area is 5 square units" almost always yields two answers, and the options usually contain both.

Illustration 2

If $C$ is the centroid of the triangle with vertices $(3,-1)$, $(1,3)$, $(2,4)$, and $P$ is the intersection of $x+3y-1=0$ and $3x-y+1=0$, which of $(9,6)$, $(-9,-6)$, $(7,6)$, $(-7,-6)$ lies on the line $CP$? (JEE previous-year question)

The centroid needs no method: $C=\left(\frac{3+1+2}{3},\frac{-1+3+4}{3}\right)=(2,2)$.

For $P$, the second equation gives $y=3x+1$; substituting into the first, $x+3(3x+1)-1=0$, so $10x+2=0$ and $x=-\frac{1}{5}$, $y=\frac{2}{5}$. Then

$m_{CP}=\frac{2-\frac{2}{5}}{2+\frac{1}{5}}=\frac{8/5}{11/5}=\frac{8}{11}$

so $CP$ is $11(y-2)=8(x-2)$, that is $8x-11y+6=0$. Testing the options, $(-9,-6)$ gives $-72+66+6=0$. Answer: $(-9,-6)$.

Illustration 3

Find the locus of the midpoints of the perpendiculars drawn from points on the line $x=2y$ to the line $x=y$. (JEE previous-year question)

Take a general point on $x=2y$ as $(2t,t)$ — again, one parameter. The foot of the perpendicular from $(x_0,y_0)$ to $x-y=0$ is $\left(\frac{x_0+y_0}{2},\frac{x_0+y_0}{2}\right)$, so here it is $\left(\frac{3t}{2},\frac{3t}{2}\right)$.

The midpoint of the segment joining $(2t,t)$ to that foot is

$(h,k)=\left(\frac{2t+\frac{3t}{2}}{2},\frac{t+\frac{3t}{2}}{2}\right)=\left(\frac{7t}{4},\frac{5t}{4}\right)$

Now eliminate $t$: dividing, $\frac{k}{h}=\frac{5}{7}$. The locus is $5x-7y=0$.

That elimination step is the whole technique for locus problems. Write the moving point's coordinates in terms of the parameter, then kill the parameter.

Illustration 4

$A(1,-1)$ and $B(0,2)$. If a point $P(x^{\prime },y^{\prime })$ makes triangle $PAB$ of area $5$ square units and lies on $3x+y-4\lambda =0$, find $\lambda$. (JEE previous-year question)

Area from the determinant form: $\frac{1}{2}\left|x^{\prime }(-1-2)+1(2-y^{\prime })+0\right|=5$, so $|-3x^{\prime }-y^{\prime }+2|=10$, giving $3x^{\prime }+y^{\prime }=12$ or $3x^{\prime }+y^{\prime }=-8$.

The point also satisfies $3x^{\prime }+y^{\prime }=4\lambda$. Matching, $4\lambda =12$ or $4\lambda =-8$, so $\lambda =3$ or $\lambda =-2$. With $3$ among the options, the answer is $\lambda =3$ — and note the second root is real, which is exactly the two-parallel-lines effect from section 6.

Part II — The circle

7. From the definition to the general form

A circle is the set of points at fixed distance $r$ from a fixed point $(h,k)$. Squaring the distance formula gives $(x-h{)}^2+(y-k{)}^2=r^2$ directly. Expanding and renaming $-h=g$, $-k=f$, $h^2+k^2-r^2=c$:

$x^2+y^2+2gx+2fy+c=0$

Read the general form backwards and you recover everything: the centre is $(-g,-f)$ — half the coefficients, negated — and $r^2=g^2+f^2-c$, so

$r=\sqrt{g^2+f^2-c}$

Three conditions must hold before an equation of this shape is a circle at all: the coefficients of $x^2$ and $y^2$ must be equal, there must be no $xy$ term, and $g^2+f^2-c$ must be positive.

When that last quantity is exactly zero the "circle" is the single point $(-g,-f)$; when it is negative there are no real points. Questions asking "for what $k$ does this represent a point" are testing exactly this, and the work is completing the square.

Two shortcuts save real time. If the endpoints of a diameter are $A(x_1,y_1)$ and $B(x_2,y_2)$, then for any point $P$ on the circle the angle $APB$ is a right angle, so $\overrightarrow{PA}\cdot \overrightarrow{PB}=0$, which is

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

And a circle touching the $x$-axis has $r=|k|$, one touching the $y$-axis has $r=|h|$, and one touching both has $|h|=|k|=r$. Reading a tangency condition straight off the picture beats setting up a discriminant.

8. Where a line meets a circle

You can substitute the line into the circle and look at the discriminant of the resulting quadratic. Do not. Compare the perpendicular distance $d$ from the centre to the line with the radius: $d<r$ gives two intersection points, $d=r$ gives tangency, $d>r$ gives none. It is one application of the distance formula against half a page of algebra.

Length of the chord cut off follows from the same picture — the perpendicular from the centre bisects the chord, so half the chord, $d$ and $r$ form a right triangle:

$\text{chord}=2\sqrt{r^2-d^2}$

The length of the tangent from an external point $P(x_1,y_1)$ comes from the same right triangle, this time with the radius as one leg: it is $\sqrt{S_1}$, where $S_1$ means the circle's expression evaluated at $P$, namely $x_1^2+y_1^2+2g{x}_1+2f{y}_1+c$. The sign of $S_1$ is also the inside/outside test — negative inside, zero on, positive outside.

9. A circle through three points

The honest method is to write $x^2+y^2+2gx+2fy+c=0$, substitute all three points, and solve three linear equations for $g,f,c$. It always works and it is not slow if you subtract pairs of equations first, which kills the quadratic constants immediately.

Take $(1,2)$, $(3,4)$, $(5,2)$. The three equations are $2g+4f+c=-5$, $6g+8f+c=-25$ and $10g+4f+c=-29$. Subtracting the first from the third eliminates both $f$ and $c$ in one step: $8g=-24$, so $g=-3$. Back-substituting gives $4f+c=1$ and $8f+c=-7$, hence $f=-2$, $c=9$. The circle is $x^2+y^2-6x-4y+9=0$, and you should confirm it by putting all three points back in — each gives zero.

The geometric route is often faster when the points are pleasant: the centre is the intersection of two perpendicular bisectors. Here the chord from $(1,2)$ to $(5,2)$ is horizontal, so its perpendicular bisector is $x=3$ on sight, giving the centre's $x$-coordinate with no algebra at all.

Illustration 5

Find the centre of the circle that passes through $(0,1)$ and touches the curve $y=x^2$ at $(2,4)$. (JEE previous-year question)

Touching a curve at a point means sharing the tangent there, which means the circle's centre lies on the curve's normal at that point. Since $y^{\prime }=2x=4$ at $x=2$, the normal has slope $-\frac{1}{4}$: $y-4=-\frac{1}{4}(x-2)$, or $x+4y=18$.

The centre is also equidistant from $(0,1)$ and $(2,4)$, both on the circle:

$h^2+(k-1{)}^2=(h-2{)}^2+(k-4{)}^2$

Expanding, the $h^2$ and $k^2$ cancel and this reduces to $4h+6k=19$. Solving with $h=18-4k$: $72-16k+6k=19$, so $k=\frac{53}{10}$ and $h=18-\frac{212}{10}=-\frac{16}{5}$. Centre $\left(-\frac{16}{5},\frac{53}{10}\right)$.

Part III — Conics: one definition, three curves

10. The definition that generates everything

Fix a point $S$ (the focus), a line (the directrix), and a positive number $e$ (the eccentricity). A conic is the set of points $P$ with

$SP=e\cdot PM$

where $PM$ is the perpendicular distance from $P$ to the directrix. That is the entire content of Part III. The parabola, ellipse and hyperbola are $e=1$, $e<1$ and $e>1$ of the same construction, and every relation you are asked to memorise falls out of it. Deriving them once is faster than memorising three tables, and far more reliable at the two-hour mark of a paper.

11. Parabola

Put $e=1$, the focus at $(a,0)$ and the directrix at $x=-a$. The definition reads $\sqrt{(x-a{)}^2+y^2}=|x+a|$. Squaring, $x^2-2ax+a^2+y^2=x^2+2ax+a^2$. The $x^2$ and $a^2$ cancel on both sides, leaving

$y^2=4ax$

Everything else is read off this. The vertex is at the origin, the axis is the $x$-axis, the focus is $(a,0)$, the directrix is $x=-a$. The latus rectum is the focal chord perpendicular to the axis: put $x=a$ to get $y=\pm 2a$, so its length is $4a$ — which is why the coefficient of $x$ is written as $4a$ in the first place.

The parametric point is $(a{t}^2,2at)$, and you should reach for it constantly. Any question about "a variable point on the parabola" becomes a question about one number $t$.

For a shifted or rotated version, the job is always to reduce to this form. Given $y^2-kx+8=0$, write it as $y^2=k\left(x-\frac{8}{k}\right)$: the vertex has moved to $\left(\frac{8}{k},0\right)$ and $4a=k$. The directrix sits $a$ behind the vertex, at $x=\frac{8}{k}-\frac{k}{4}$. Setting that equal to a given directrix turns the geometry into one quadratic in $k$.

Illustration 6

The line $x=1$ is the directrix of the parabola $y^2-kx+8=0$. Find $k$. (JEE previous-year question)

Put the parabola in standard shape: $y^2=kx-8=k\left(x-\frac{8}{k}\right)$. So the vertex is at $\left(\frac{8}{k},0\right)$ and $4a=k$, giving $a=\frac{k}{4}$.

The directrix lies a distance $a$ behind the vertex, at $x=\frac{8}{k}-\frac{k}{4}=1$. Multiplying by $4k$: $32-k^2=4k$, so $k^2+4k-32=0$ and $k=4$ or $k=-8$. Both are valid — one opens right, the other left — so check which appears in the options.

Illustration 7

The locus of the midpoint of the segment joining the focus of $y^2=4ax$ to a moving point on the parabola is another parabola. Find its directrix. (JEE previous-year question)

The moving point is $(a{t}^2,2at)$ and the focus is $(a,0)$, so the midpoint is

$(h,k)=\left(\frac{a+a{t}^2}{2},at\right)$

From the second coordinate $t=\frac{k}{a}$; substituting into the first, $2ah=a^2+k^2$, that is $k^2=2a\left(h-\frac{a}{2}\right)$.

This is a parabola with vertex $\left(\frac{a}{2},0\right)$ and $4A=2a$, so $A=\frac{a}{2}$. Its directrix is $A$ behind the vertex: $x=\frac{a}{2}-\frac{a}{2}$. The directrix is $x=0$, the $y$-axis.

12. Ellipse

Now take $e<1$, focus $(ae,0)$ and directrix $x=a/e$. The definition gives $(x-ae{)}^2+y^2=e^2{\left(x-\frac{a}{e}\right)}^2$. Expand the right side: $e^2{x}^2-2aex+a^2$. Expand the left: $x^2-2aex+a^2{e}^2+y^2$. The $-2aex$ terms cancel on both sides, and collecting what is left gives $x^2(1-e^2)+y^2=a^2(1-e^2)$. Dividing through by $a^2(1-e^2)$:

$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$

Define $b^2=a^2(1-e^2)$ and it is the standard ellipse. Notice that this relation was produced by the derivation rather than asserted — that is the point of doing it. Since $e<1$, $b<a$ automatically.

So: semi-major axis $a$, semi-minor $b$, foci at $(\pm ae,0)$, directrices at $x=\pm a/e$, latus rectum $2b^2/a$, and the sum of the two focal distances of any point is $2a$. The parametric point is $(a\cos \theta ,b\sin \theta )$, which is the circle's parametrisation stretched — and that stretch is exactly what the auxiliary circle $x^2+y^2=a^2$ records.

One warning that costs marks every year: when $b>a$ in the equation you are handed, the major axis is vertical, the foci are at $(0,\pm be)$, and the eccentricity is $e^2=1-a^2/b^2$. Always compare the two denominators before writing anything down. In $25(x+1{)}^2+9(y+2{)}^2=225$, dividing by $225$ gives $\frac{(x+1{)}^2}{9}+\frac{(y+2{)}^2}{25}=1$; the larger denominator is under $y$, so the major axis is vertical, $c^2=25-9=16$, and the foci are $4$ above and below the centre $(-1,-2)$ — at $(-1,2)$ and $(-1,-6)$.

Illustration 8

$B$ is an end of the minor axis of an ellipse and $F,F^{\prime }$ are its foci. If $\angle FB{F}^{\prime }=90^{\circ }$, find the eccentricity. (JEE previous-year question)

Put $B=(0,b)$, $F=(ae,0)$, $F^{\prime }=(-ae,0)$. The right angle at $B$ means $\overrightarrow{BF}\cdot \overrightarrow{B{F}^{\prime }}=0$:

$(ae)(-ae)+(-b)(-b)=0⟹b^2=a^2{e}^2$

But $b^2=a^2(1-e^2)$ always. Equating, $a^2{e}^2=a^2(1-e^2)$, so $2e^2=1$. Answer: $e=\frac{1}{\sqrt{2}}$.

Notice the shape of that argument — one geometric condition, one standing identity, one unknown. Most conic questions are exactly this.

13. Hyperbola

Run the identical derivation with $e>1$. Every line is the same until the last step, where $1-e^2$ is now negative. Writing $b^2=a^2(e^2-1)$ to keep $b^2$ positive:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

That is the only difference between the two curves, and it is the source of every sign you have to get right. The ellipse has $b^2=a^2(1-e^2)$; the hyperbola has $b^2=a^2(e^2-1)$. Rearranged: $e^2=1-b^2/a^2$ for the ellipse, $e^2=1+b^2/a^2$ for the hyperbola. If you ever cannot remember which takes the plus, recall that a hyperbola must have $e>1$, so its formula must be the one that adds.

Foci $(\pm ae,0)$, latus rectum $2b^2/a$ — both identical in form to the ellipse. What is new is the asymptotes, $y=\pm \frac{b}{a}x$, the lines the curve approaches at infinity. They come from setting the right-hand side to $0$ instead of $1$: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ factors as $\left(\frac{x}{a}-\frac{y}{b}\right)\left(\frac{x}{a}+\frac{y}{b}\right)=0$, a pair of straight lines. The parametric point is $(a\sec \theta ,b\tan \theta )$, chosen because ${\sec }^2\theta -{\tan }^2\theta =1$ matches the equation's minus sign.

Illustration 9

For a hyperbola the latus rectum is $8$ and the conjugate axis equals half the distance between the foci. Find the eccentricity. (JEE previous-year question)

Translate both statements. Conjugate axis $=2b$; distance between foci $=2ae$; so $2b=ae$, giving $b=\frac{ae}{2}$ and $b^2=\frac{a^2{e}^2}{4}$.

Combine with the standing identity $b^2=a^2(e^2-1)$: $a^2(e^2-1)=\frac{a^2{e}^2}{4}$, so $e^2-1=\frac{e^2}{4}$ and $\frac{3e^2}{4}=1$. Answer: $e=\frac{2}{\sqrt{3}}$.

The latus rectum was not needed for the eccentricity — it only fixes the size. If you want it, $\frac{2b^2}{a}=8$ with $b^2=a^2(e^2-1)=\frac{a^2}{3}$ gives $a=12$. Recognising that a datum is redundant is itself worth seconds.

Illustration 10

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, find $b^2$. (JEE previous-year question)

The hyperbola is not in standard form — the right side must be $1$. Dividing, $\frac{x^2}{144/25}-\frac{y^2}{81/25}=1$, so $A^2=\frac{144}{25}$ and $B^2=\frac{81}{25}$.

Its eccentricity: $E^2=1+\frac{81}{144}=\frac{225}{144}$, so $E=\frac{5}{4}$ and the foci are at $\pm AE=\pm \frac{12}{5}\cdot \frac{5}{4}=\pm 3$.

For the ellipse the foci are at $\pm \sqrt{16-b^2}$, so $16-b^2=9$. Answer: $b^2=7$.

14. Reading a conic off a messy equation

Faced with $A{x}^2+B{y}^2+Cx+Dy+E=0$ and no $xy$ term, complete the square in $x$ and in $y$ separately, then divide to make the right side $1$. What you get tells you the curve: same signs on both squared terms is an ellipse (a circle if the coefficients are equal), opposite signs a hyperbola, one squared term missing a parabola.

Try $x^2+2y^2-2x+3y+2=0$. Completing squares, $(x-1{)}^2-1+2{\left(y+\frac{3}{4}\right)}^2-\frac{9}{8}+2=0$, so $(x-1{)}^2+2{\left(y+\frac{3}{4}\right)}^2=\frac{1}{8}$. Dividing, $\frac{(x-1{)}^2}{1/8}+\frac{(y+3/4{)}^2}{1/16}=1$. Both signs positive, so it is an ellipse; the larger denominator is under $x$, so $a^2=\frac{1}{8}$ and $b^2=\frac{1}{16}$, giving $e^2=1-\frac{1/16}{1/8}=\frac{1}{2}$ and $e=\frac{1}{\sqrt{2}}$. The messy fractions never mattered — only the ratio did.

Illustration 11

Show that the curves $4x^2+p{y}^2=45$ and $x^2-4y^2=5$ cut orthogonally, and find $p$. (JEE previous-year question)

Orthogonal intersection means the tangents at each common point are perpendicular, so the product of the two gradients there is $-1$.

Differentiating implicitly: from the first, $8x+2py{y}^{\prime }=0$, so $y^{\prime }=-\frac{4x}{py}$. From the second, $2x-8y{y}^{\prime }=0$, so $y^{\prime }=\frac{x}{4y}$. Their product must be $-1$:

$-\frac{4x}{py}\cdot \frac{x}{4y}=-1⟹x^2=p{y}^2$

Now use the two curve equations at that point. Substituting into the first, $4p{y}^2+p{y}^2=45$, so $p{y}^2=9$. Substituting into the second, $p{y}^2-4y^2=5$, so $9-4y^2=5$ and $y^2=1$. Hence $p=9$.

Part IV — Tangents and normals: one calculation, four results

15. The tangency condition, derived once

Most candidates memorise four separate conditions for $y=mx+c$ to touch a curve. Three of them are the same calculation.

Write any central conic as $\frac{x^2}{A}+\frac{y^2}{B}=1$. The circle $x^2+y^2=a^2$ is $A=B=a^2$; the ellipse is $A=a^2,B=b^2$; the hyperbola is $A=a^2,B=-b^2$. Substituting $y=mx+c$:

$\frac{x^2}{A}+\frac{(mx+c{)}^2}{B}=1⟹x^2(B+A{m}^2)+2Amcx+A(c^2-B)=0$

Tangency means this quadratic has equal roots, so its discriminant vanishes: $4A^2{m}^2{c}^2=4A(B+A{m}^2)(c^2-B)$. Expanding the right side gives $B{c}^2-B^2+A{m}^2{c}^2-AB{m}^2$, all multiplied by $A$. The $A{m}^2{c}^2$ terms cancel against the left, leaving $B{c}^2=B^2+AB{m}^2$, and dividing by $B$:

$c^2=A{m}^2+B$

That one line is three memorised results. For the circle, $c^2=a^2{m}^2+a^2=a^2(1+m^2)$. For the ellipse, $c^2=a^2{m}^2+b^2$. For the hyperbola, put $B=-b^2$ and get $c^2=a^2{m}^2-b^2$. The famous sign difference between the last two is not a fourth fact to store; it is the same $B$ carrying its own sign.

The parabola is not central, so it needs its own line of work. Substituting $y=mx+c$ into $y^2=4ax$ gives $m^2{x}^2+(2mc-4a)x+c^2=0$, and equal roots require $(2mc-4a{)}^2=4m^2{c}^2$. The $4m^2{c}^2$ cancels, leaving $-16amc+16a^2=0$:

$c=\frac{a}{m}\text{so the tangent is }y=mx+\frac{a}{m}$

Illustration 12

If $3x+4y=12\sqrt{2}$ is a tangent to $\frac{x^2}{a^2}+\frac{y^2}{9}=1$, find the distance between the foci. (JEE previous-year question)

Rewrite the line as $y=-\frac{3}{4}x+3\sqrt{2}$, so $m=-\frac{3}{4}$ and $c=3\sqrt{2}$. The tangency condition $c^2=a^2{m}^2+b^2$ gives $18=\frac{9a^2}{16}+9$, hence $a^2=16$.

Then $e^2=1-\frac{9}{16}=\frac{7}{16}$ and $e=\frac{\sqrt{7}}{4}$, so the distance between the foci is $2ae=2\cdot 4\cdot \frac{\sqrt{7}}{4}$. Answer: $2\sqrt{7}$.

Illustration 13

Find the relation between the slopes of the tangents to $\frac{x^2}{25}-\frac{y^2}{16}=1$ drawn from the point $(6,2)$. (JEE previous-year question)

Any tangent is $y=mx+c$ with $c^2=25m^2-16$. Passing through $(6,2)$ forces $c=2-6m$, so

$(2-6m{)}^2=25m^2-16⟹4-24m+36m^2=25m^2-16⟹11m^2-24m+20=0$

Hence $m_1+m_2=\frac{24}{11}$ and $m_1{m}_2=\frac{20}{11}$. The general move — impose the tangency condition, then impose passing through the point, and read the two slopes off one quadratic — solves every "pair of tangents from a point" question, including the director-circle case where you would set $m_1{m}_2=-1$.

Illustration 14

Find the equations of the common tangents to the circle $x^2+y^2=2a^2$ and the parabola $y^2=8ax$, and the area they enclose with the two chords of contact when $a=1$. (JEE previous-year question)

A tangent to $y^2=8ax$ (so $4A=8a$, $A=2a$) is $y=mx+\frac{2a}{m}$. For it to touch the circle, its distance from the origin must equal $a\sqrt{2}$:

$\frac{|2a/m|}{\sqrt{1+m^2}}=a\sqrt{2}⟹\frac{4a^2}{m^2}=2a^2(1+m^2)⟹m^4+m^2-2=0$

Factoring, $(m^2+2)(m^2-1)=0$, and only $m=\pm 1$ is real. The common tangents are $y=\pm (x+2a)$.

For the area with $a=1$: the contact points on the parabola are $\left(\frac{A}{m^2},\frac{2A}{m}\right)=(2,\pm 4)$, and on the circle they are the feet of the perpendiculars from the origin, $(-1,\pm 1)$. Those four points form a trapezium with parallel vertical sides of length $8$ and $2$, three units apart, so the area is $\frac{1}{2}(8+2)(3)=$ 15 square units.

16. Tangent at a known point: where the T = 0 rule comes from

If the point of contact is known rather than the slope, differentiate. On the ellipse, $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$, so at $(x_1,y_1)$ the slope is $-\frac{b^2{x}_1}{a^2{y}_1}$. The tangent is $y-y_1=-\frac{b^2{x}_1}{a^2{y}_1}(x-x_1)$, which clears to $b^2{x}_{1}x+a^2{y}_{1}y=b^2{x}_1^2+a^2{y}_1^2$. But $(x_1,y_1)$ is on the ellipse, so $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1$, meaning that right-hand side is exactly $a^2{b}^2$. Dividing through:

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

Compare that with the original equation and the pattern is plain: $x^2$ became $x{x}_1$ and $y^2$ became $y{y}_1$. The same substitution works on every conic, and the full rule — usually written $T=0$ — replaces $x^2\to x{x}_1$, $y^2\to y{y}_1$, $x\to \frac{x+x_1}{2}$, $y\to \frac{y+y_1}{2}$ and $xy\to \frac{x{y}_1+x_{1}y}{2}$. For the parabola $y^2=4ax$ it gives $y{y}_1=2a(x+x_1)$; for the circle $x^2+y^2=a^2$ it gives $x{x}_1+y{y}_1=a^2$.

The normal is then just the perpendicular through the same point. For the ellipse this tidies into $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$, and for the hyperbola into $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$.

Illustration 15

A hyperbola has vertices $(\pm 6,0)$ and passes through $P(10,16)$. Find the equation of the normal at $P$. (JEE previous-year question)

Vertices at $(\pm 6,0)$ give $a=6$. Substituting $P$: $\frac{100}{36}-\frac{256}{b^2}=1$, so $\frac{256}{b^2}=\frac{64}{36}$ and $b^2=144$.

The hyperbola normal at $(x_1,y_1)$ is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$, which here reads $\frac{36x}{10}+\frac{144y}{16}=180$, that is $3.6x+9y=180$. Dividing by $1.8$: $2x+5y=100$.

17. The parabola's normal-chord relation

This one is worth deriving because it is asked directly. Take the point $t_1$, that is $(a{t}_1^2,2a{t}_1)$. From $y^2=4ax$, $2y{y}^{\prime }=4a$, so $y^{\prime }=\frac{2a}{y}=\frac{1}{t_1}$ and the normal has slope $-t_1$. Its equation is $y-2a{t}_1=-t_1(x-a{t}_1^2)$, that is $y=-t_{1}x+a{t}_1^3+2a{t}_1$.

Ask where it meets the parabola again, at the point $t_2$. Substituting $x=a{t}_2^2$, $y=2a{t}_2$ gives $2a{t}_2=-a{t}_1{t}_2^2+a{t}_1^3+2a{t}_1$. Divide by $a$ and gather: $2(t_2-t_1)=-t_1(t_2^2-t_1^2)=-t_1(t_2-t_1)(t_2+t_1)$. The two points are distinct so $t_2\ne t_1$, and cancelling that factor leaves $2=-t_1(t_1+t_2)$:

$t_2=-t_1-\frac{2}{t_1}$

18. Director circle: perpendicular tangents

From which points do the two tangents to a conic meet at right angles? Let the point be $(h,k)$. A tangent through it has $c=k-mh$, and by the condition above $c^2=A{m}^2+B$, so $(k-mh{)}^2=A{m}^2+B$. Expanding and collecting in $m$:

$m^2(h^2-A)-2hkm+(k^2-B)=0$

The two roots are the two tangent slopes. Perpendicular means their product is $-1$, and the product of the roots is $\frac{k^2-B}{h^2-A}$. Setting that to $-1$ gives $k^2-B=A-h^2$:

$x^2+y^2=A+B$

One derivation, three answers. The circle $x^2+y^2=a^2$ has director circle $x^2+y^2=2a^2$; the ellipse gives $x^2+y^2=a^2+b^2$; the hyperbola, with $B=-b^2$, gives $x^2+y^2=a^2-b^2$. For a hyperbola with $b>a$ that circle is imaginary, which is the correct answer, not an error: no point sees such a hyperbola at a right angle.

Summary

Coordinate geometry converts a picture into equations, and the marks are in choosing the conversion, not performing it. Three habits do most of the work.

Parametrise before you minimise. A point on a curve should be one unknown, not two: $(a{t}^2,2at)$ on a parabola, $(a\cos \theta ,b\sin \theta )$ on an ellipse, $(a\sec \theta ,b\tan \theta )$ on a hyperbola. The opening problem collapsed from a constrained two-variable minimisation to a quadratic because of this alone, and every locus question is the same move followed by eliminating the parameter.

Derive the conic relations, do not memorise them. The focus–directrix definition $SP=e\cdot PM$ produces $y^2=4ax$ at $e=1$, the ellipse at $e<1$ and the hyperbola at $e>1$, and it produces $b^2=a^2(1-e^2)$ and $b^2=a^2(e^2-1)$ as by-products rather than as facts to store. The whole sign confusion between ellipse and hyperbola lives at one place: whether $1-e^2$ or $e^2-1$ is the positive one.

Learn the conditions in their unified form. Writing a central conic as $\frac{x^2}{A}+\frac{y^2}{B}=1$, one discriminant calculation yields $c^2=A{m}^2+B$ for tangency and one more yields $x^2+y^2=A+B$ for the director circle — covering circle, ellipse and hyperbola at once, with the sign of $B$ doing all the work. The parabola needs its own line, $c=a/m$, because it is not central.

Beyond that, four small results pay for themselves repeatedly: compare distance-to-centre with radius rather than substituting a line into a circle; use $L_1+\lambda L_2=0$ instead of solving for an intersection; remember that a homogeneous quadratic in $x$ and $y$ is a pair of lines through the origin; and treat "equal areas from an interior point" as a synonym for "centroid".