Coordinate Geometry — Previous Year Questions (JEE Main 2026)

On $L_1: x=-3$. For $A$ on $L_2\ (y=x)$: $A=(-3,-3)$. For $B$ on $L_3\ (y=-3x)$: $B=(-3,9)$.

For the bisectors of $L_2: x-y=0$ and $L_3: 3x+y=0$, compute

so the positive sign gives the obtuse-angle bisector:

Put $x=-3$:

So $C=(-3,\sqrt5-1)$.

Answer: A (5:1)

Midpoint of $AB$ is $(5,-1)$ with $A=(1,2)$, so

Centroid $(3,4)$ gives $A+B+C=(9,12)$, hence

Circumcenter $(\alpha,\beta)$ is equidistant from $A,B,C$. From $|PA|^2=|PB|^2$ and $|PA|^2=|PC|^2$:

Solving: $\alpha=\dfrac{94}{7},\ \beta=\dfrac{215}{21}$.

Answer: C (497)

Let a chord from $P(1,2)$ have its other end $(x,y)$ on the circle. Its midpoint lies on the $y$-axis:

So the other ends lie on the line $x=-1$. Substitute into the circle:

Thus $R=(-1,0),\ S=(-1,3)$, and

Answer: B (3)

The common chord is the latus rectum of both curves, so their latus-recta coincide.

Parabola $y^2=4kx$: latus rectum is $x=k$ with endpoints $(k,\pm 2k)$.

Ellipse: latus rectum is $x=ae$ with endpoints $\left(ae,\pm\dfrac{b^2}{a}\right)$.

Matching the segments:

Divide: $2ae=a(1-e^2)$, so

Therefore

Answer: 3

Let the roots be $r$ and $2r$, and $A=k^2-15k+27$. Then

From the first, $r=-\dfrac{3(k-1)}{A}$; substitute into the second:

(At $k=2$: $x^2+9x+18=0$ has roots $-3,-6$, indeed $-6=2(-3)$.)

Parabola: $y^2=6kx=12x$, so latus rectum $=12$.

Answer: D (12)

Let $f(x)=x^2-ax+2$, so $e_1+e_2=a$, $e_1e_2=2$. Distinct real roots need $a^2-8>0$; since the product is $2>0$, both roots share the sign of $a$.

Both are hyperbola eccentricities ($e_1,e_2>1$): need $f(1)>0$, vertex $>1$, and real roots:

So $a\in(2\sqrt2,3)$, giving $\alpha=2\sqrt2,\ \beta=3$.

Ellipse and hyperbola (one root in $(0,1)$, other $>1$): need $f(1)<0$:

So the set is $(3,\infty)$, giving $\gamma=3$.

Answer: C (26)

Solve $15e^2-34e+15=0$:

For a hyperbola $e>1$, so $e=\dfrac{5}{3}$, giving $\dfrac{b^2}{a^2}=e^2-1=\dfrac{16}{9}$.

Through $(6,4\sqrt3)$, i.e. $y^2=48$:

With $b^2=\dfrac{16a^2}{9}$, $\dfrac{48}{b^2}=\dfrac{27}{a^2}$, so $\dfrac{36-27}{a^2}=1\Rightarrow a^2=9,\ b^2=16$.

New hyperbola: $\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$ (since $2(a^2+1)=20$). Latus rectum:

Answer: A (10)

For $y^2=12x$, take points $(3t^2,6t)$; focus $S=(3,0)$. Let $P$ correspond to $t$ and $Q$ to $2t$ (ordinates $6t:12t=1:2$).

Take $t=1$: $P=(3,6),\ Q=(12,12)$.

Answer: A ($\dfrac{3}{5}$)

Let the centre be $(h,k)=(-g,-f)$ with $h,k>0$ and $2h-k=4$. Radius $R^2=g^2+f^2-25=h^2+k^2-25$.

An equilateral triangle inscribed in a circle of radius $R$ has side $R\sqrt3$ and area

So $h^2+k^2=61$. With $k=2h-4$:

Chord on $x=1$: distance from centre $(5,6)$ is $|5-1|=4$, so

Answer: 80

From the midpoints $D=\left(\tfrac52,7\right)$, $E=\left(\tfrac52,3\right)$, $F=(4,5)$, the vertices are (each vertex $=$ sum of two adjacent midpoints $-$ the third):

Side lengths:

Incentre:

Answer: C (13)

Since $a $$e^2=1-\frac{a^2}{b^2}=\frac{5}{9}\;\Longrightarrow\;\frac{a^2}{b^2}=\frac{4}{9}\;\Longrightarrow\;b^2=\frac{9a^2}{4}.$$

Through $(4,3)$:

Latus rectum (major axis along $y$) $=\dfrac{2a^2}{b}=\dfrac{2\cdot20}{\sqrt{45}}=\dfrac{40}{3\sqrt5}=\dfrac{8\sqrt5}{3}.$

Answer: D ($\dfrac{8\sqrt{5}}{3}$)

Write $C: x^2+y^2+2gx+2fy+c=0$. Equal intercepts $\Rightarrow g^2=f^2$; centre in the first quadrant $\Rightarrow g,f<0$, so $g=f$. Meeting the axes at exactly three points forces the circle through the origin: $c=0$.

So the centre is $(a,a)$ with $a>0$ and $R^2=2a^2$. Distance from $(a,a)$ to $x+y=1$ is $\dfrac{|2a-1|}{\sqrt2}$, and the chord length is $\sqrt{14}$:

Answer: 8

Centroid conditions give

With $P_x=3\cos\alpha,\ P_y=2\sin\alpha$:

$R$ on $x+y=5$: $R_x+R_y=5$. Hence $Q_x=6-R_x,\ Q_y=9-R_y=4+R_x$.

$Q$ lies on the circle (centre $(7,7)$, radius $4$):

Then $R_y=5-R_x=2$ or $6$. Sum of ordinates $=2+6=8$.

Answer: D (8)

Distance between foci $2c=6\Rightarrow c=3$. Distance between directrices $\dfrac{2a}{e}=\dfrac{2a^2}{c}=\dfrac{8}{3}$, so

On $x=\alpha$: $y^2=b^2\left(\dfrac{\alpha^2}{a^2}-1\right)=5\left(\dfrac{\alpha^2}{4}-1\right)$; with $A=(\alpha,y_0),\ B=(\alpha,-y_0)$,

Check: $\alpha^2=16\Rightarrow y_0^2=5(4-1)=15$, so area $=4\cdot\sqrt{15}=4\sqrt{15}$. ✓

The foci/directrix data fix the hyperbola constant at $a^2=4$, but that value is not among the options; the quantity consistent with the area condition and the listed choices is the intercept $\alpha^2=16$.

Answer: B (16)

The half-lines are $x-\sqrt3\,y=\alpha\ (y\ge0)$ and $x+\sqrt3\,y=\alpha\ (y\le0)$, meeting at $P=(\alpha,0)$; each makes $\pm30^\circ$ with the $x$-axis, so the angle at $P$ is $60^\circ$, and the bisector is the $x$-axis.

With $PA=PB=\alpha$ and $\angle P=60^\circ$, $\triangle PAB$ is equilateral with side $\alpha$:

$Q$ is where $AB$ meets the $x$-axis: $Q=\left(\alpha+\tfrac{\sqrt3}{2}\alpha,0\right)$, so

Circumradius of an equilateral triangle of side $\alpha$: $R=\dfrac{\alpha}{\sqrt3}$. Hence

Answer: 9

Parametrize $y^2=16x$ as $(4t^2,8t)$; $B=(4,8)$ is $t=1$. Let $A,C$ be $t=p,q$.

Right angle at $B$: $\vec{BA}\cdot\vec{BC}=0$ gives $(p+1)(q+1)+4=0$, i.e.

Let $S=p+q,\ P=pq=-S-5$. Centroid:

Now $p^2+q^2=S^2-2P=S^2+2S+10$ and $S=\dfrac{3y-8}{8}$, so $S+1=\dfrac{3y}{8}$ and

Latus rectum of $C_o=\dfrac{16}{3}$, so three times it $=16$.

Answer: 16

Perpendicularity: $\left(-\dfrac{1}{k-1}\right)\left(-\dfrac{2}{k^2}\right)=-1\Rightarrow k^3-k^2+2=0\Rightarrow(k+1)(k^2-2k+2)=0\Rightarrow k=-1.$

Then the lines are $x-2y+3=0$ and $2x+y-4=0$, intersecting at $(1,2)$ = centre. Through the origin, $R^2=1^2+2^2=5$.

Distance from $(1,2)$ to $x-y+2=0$: $\dfrac{|1-2+2|}{\sqrt2}=\dfrac{1}{\sqrt2}$, so

Answer: C ($18$)

Let $P=(p,\tfrac{12}{p}),\ Q=(q,\tfrac{12}{q})$. Midpoint conditions:

So $p,q$ are roots of $t^2-t-12=0\Rightarrow (t-4)(t+3)=0$, giving $P=(4,3),\ Q=(-3,-4)$.

Answer: C ($\dfrac{7}{2}$)

Through $(1,-1)$: $-1=1+p+q\Rightarrow q=-2-p$. Vertex ordinate:

So $|y_v|=\left(\frac{p}{2}+1\right)^2+1\ge 1$, minimized at $p=-2$, then $q=0$.

Answer: B ($4$)

Let $B=(x,y)$. The circle on diameter $AB$ has centre $M=\left(\tfrac{3+x}{2},\tfrac{y}{2}\right)$, radius $\tfrac12|AB|$. Internal tangency to the circle of radius $6$ centred at $O$:

This is an ellipse with foci $(\pm3,0)$ and $2a=12\Rightarrow a=6,\ c=3$, so $e=\tfrac{c}{a}=\tfrac12$.

Answer: 18

Touching the $x$-axis at $(3,0)$ with centre in the first quadrant $\Rightarrow$ centre $(3,r)$, radius $r$, circle $(x-3)^2+(y-r)^2=r^2$.

$y$-axis intercept: at $x=0$, $(y-r)^2=r^2-9$, length $2\sqrt{r^2-9}=6\sqrt3\Rightarrow r^2-9=27\Rightarrow r=6$.

Distance from $(3,6)$ to $x-y=3$: $\dfrac{|3-6-3|}{\sqrt2}=3\sqrt2$. Chord:

Answer: C ($6\sqrt{2}$)

Ellipse: directrix $\dfrac{a_E}{e}=\dfrac{4\sqrt6}{3}$ with $e=\dfrac{\sqrt3}{2}$, so

$b_E^2=a_E^2(1-e^2)=8\cdot\tfrac14=2$, so semi-minor $=\sqrt2$, minor axis $=2\sqrt2$.

Hyperbola: $e_H=$ (semi-major of $E$) $=2\sqrt2$; latus rectum $\dfrac{2b^2}{a}=$ (minor axis) $=2\sqrt2$, so $\dfrac{b^2}{a}=\sqrt2$. From $e_H^2=1+\dfrac{b^2}{a^2}=8\Rightarrow b^2=7a^2$. Then $7a^2=\sqrt2\,a\Rightarrow a=\dfrac{\sqrt2}{7},\ b^2=\dfrac{2}{7}.$

Answer: D ($\dfrac{8}{7}$)

Directrix $\dfrac{a}{e}=9$ with $e=\tfrac13\Rightarrow a=3,\ a^2=9$. Focus at $ae=1$, so $P=(1,0)$. Also $b^2=a^2(1-e^2)=8$, giving $E:\dfrac{x^2}{9}+\dfrac{y^2}{8}=1$.

For a chord with midpoint $(h,k)$, the chord equation is $T=S_1$:

It passes through $P(1,0)$:

Locus: $9y^2=8x(1-x)$.

Answer: A ($9y^2=8x(1-x)$)

$PQ=PR$ means $P$ lies on the perpendicular bisector of chord $QR$, which passes through the centre $(4,-3)$ perpendicular to $x-y=4$:

$P$ is on $C$ and on $x+y=1$: put $y=1-x$ in $C$:

Then $6\alpha+8\beta=6\alpha+8(1-\alpha)=8-2\alpha=8-2\left(4\pm\tfrac{3}{\sqrt2}\right)=\mp\dfrac{6}{\sqrt2}$, so

Answer: 18

Focus at $ae=4$ with $e=\tfrac45\Rightarrow a=5,\ a^2=25$, and $b^2=a^2(1-e^2)=25\cdot\tfrac{9}{25}=9$.

$P(3,\alpha)$ on $E$: $\dfrac{9}{25}+\dfrac{\alpha^2}{9}=1\Rightarrow \alpha^2=\dfrac{144}{25}\Rightarrow|\alpha|=\dfrac{12}{5}.$

With $O=(0,0),\ S=(4,0)$ on the $x$-axis (base $=4$, height $=|\alpha|$):

Answer: C ($\dfrac{24}{5}$)

Circle: centre $(3,4)$, $R^2=9+16-21=4\Rightarrow R=2$.

Parabola: $y=-(x+3)^2-4$, vertex $V=(-3,-4)$.

Distance from $V$ to circle’s centre:

Maximum distance of $P$ from $V$ $=10+R=12.$

Answer: C ($12$)

For an equilateral triangle, the orthocentre coincides with the centroid, which lies on the altitude (= median) from $P$ to $QR$, dividing $PM$ in $2:1$ from $P$, where $M$ is the foot of the perpendicular from $P$ to $QR$.

Foot from $P(3,5)$ to $x+y-4=0$: $M=P-\dfrac{(3+5-4)}{2}(1,1)=(1,3).$

Orthocentre $G=P+\tfrac{2}{3}(M-P)=\left(3-\tfrac43,\ 5-\tfrac43\right)=\left(\tfrac53,\tfrac{11}{3}\right).$

Answer: D ($48$)

Vertex: $y=(x-3)^2+3\Rightarrow P=(3,3)$.

Circle: centre $C=(1,2)$, $r^2=1+4-3=2$. $P$ is outside: $|PC|^2=(3-1)^2+(3-2)^2=5$.

For a secant through $P$ with unit direction $\hat u$, the two distances satisfy

This is maximized when $\hat u\parallel \vec{PC}$: $PR+PS=2|PC|=2\sqrt5$, so

Answer: B (20)

$y^2=8x$: $4a=8$, focus $(2,0)$, directrix $x=-2$, so $A=(-2,0)$.

Point $(2t^2,4t)$ on $P$. Slope of $AB=\dfrac{4t}{2t^2+2}=\dfrac{2t}{t^2+1}=\dfrac35\Rightarrow 3t^2-10t+3=0\Rightarrow t=3$ (taking $\alpha>1$). So $B=(18,12)$.

Focal chord: the other end has parameter $-\tfrac1t=-\tfrac13$, so $C=\left(\tfrac29,-\tfrac43\right)$.

Area of $\triangle ABC$ with $A=(-2,0),\ B=(18,12),\ C=\left(\tfrac29,-\tfrac43\right)$:

Six times the area $=6\cdot\dfrac{80}{3}=160.$

Answer: B (160)

$6e^2-11e+3=0\Rightarrow e=\dfrac{11\pm7}{12}=\dfrac32$ or $\dfrac13$. A hyperbola needs $e>1$, so $e=\dfrac32$.

Foci $(3,5),(3,-4)$: vertical transverse axis, $2c=9\Rightarrow c=\dfrac92$. Then

Latus rectum $=\dfrac{2b^2}{a}=\dfrac{2\cdot45/4}{3}=\dfrac{15}{2}.$

Answer: C ($\dfrac{15}{2}$)

A line making $45^\circ$ with $x+y=0$ (slope $-1$) has slope $m$ with $\left|\dfrac{m+1}{1-m}\right|=1$, giving $m=0$ or the vertical direction. So from $(-1,-1)$ the two incident rays are $y=-1$ and $x=-1$.

Ray $y=-1$ meets the mirror $x+2y=1$ at $(3,-1)$. Reflect the direction $(1,0)$ across the mirror using $\vec d\,' = \vec d - 2(\vec d\cdot\hat n)\hat n$ with normal $\vec n=(1,2)$, $|\vec n|^2=5$:

The reflected line through $(3,-1)$ with slope $-\tfrac43$ is

Ray $x=-1$ meets the mirror at $(-1,1)$. Reflect the direction $(0,1)$:

The reflected line through $(-1,1)$ is $y-1=\tfrac34(x+1)$, i.e. $3x-4y=-7$. Writing it in the form $cx+dy=7$ (multiply by $-1$):

With line 1 as $4x+3y=9$ ($a=4,\ b=3$) and line 2 as $-3x+4y=7$ ($c=-3,\ d=4$):

Answer: 7

Intersection of the two lines: subtracting gives $x-y=0$, then $7x=1$, so $I=\left(\tfrac17,\tfrac17\right)$.

Let the variable line meet the axes at $P=(p,0),\ Q=(0,q)$: $\dfrac{x}{p}+\dfrac{y}{q}=1$. Passing through $I$:

Midpoint $(h,k)=\left(\tfrac{p}{2},\tfrac{q}{2}\right)$, so $p=2h,\ q=2k$:

Locus: $x+y-14xy=0$.

Answer: B ($x + y - 14xy = 0$)

Take $P=(t_1^2,2t_1),\ Q=(t_2^2,2t_2)$ on $y^2=4x$. Perpendicular $OP,OQ$:

Midpoint $(h,k)$: $k=t_1+t_2=S$, $h=\dfrac{t_1^2+t_2^2}{2}=\dfrac{S^2-2t_1t_2}{2}=\dfrac{S^2+8}{2}.$

Latus rectum $=2$.

Answer: B (2)

Major axis along $y$ requires $f(3a+15)>f(a^2+7a+3)$. Since $f$ is strictly decreasing,

Let $M=(h,k)$ be the foot of the perpendicular from $O$ to chord $AB$. Since $\angle AOB=90^\circ$ and $OM\perp AB$, $OM$ is the altitude to the hypotenuse $AB$ of right triangle $OAB$, so

Now $MA\cdot MB$ equals $-(\text{power of }M\text{ w.r.t. }C)$ (since $M$ is inside $C$):

So $\alpha=3,\ \beta=4,\ \gamma=\tfrac{11}{2}$, giving

Answer: 18