Ellipse JEE Main previous year questions with solutions

5 solved JEE Main questions on Ellipse, free to read — no sign-in needed. The full chapter has 95 questions; sign in to attempt the remaining 90 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Equation of Ellipse
   Let $\dfrac{x^2}{f(a^2+7a+3)} + \dfrac{y^2}{f(3a+15)} = 1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of $a$ is $\mathbb{R} - [\alpha, \beta]$, then $\alpha^2+\beta^2$ is equal to:
   1. A.$28$
   2. B.$40$
   3. C.$61$
   4. D.$24$
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   Answer: (B)

   For the given equation to represent an ellipse with its major axis along the $y$-axis, the denominator of $y^2$ must be strictly greater than the denominator of $x^2$. $f(3a+15) \gt f(a^2+7a+3)$ Since $f$ is given as a strictly decreasing positive function on $\mathbb{R}$, the inequality sign reverses for the arguments: $3a+15 \lt a^2+7a+3$ $a^2 + 4a - 12 \gt 0$ $(a+6)(a-2) \gt 0$ The solution to this inequality is $a \in (-\infty, -6) \cup (2, \infty)$. This can be rewritten in terms of the set difference as $a \in \mathbb{R} - [-6, 2]$. Comparing this with the given set $\mathbb{R} - [\alpha, \beta]$, we get $\alpha = -6$ and $\beta = 2$. $\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$ Answer: $40$
2. Q2JEE Main 2025 (24 Jan, Shift 2)Chord with given Middle Point
   The equation of the chord, of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid-point is $(3,1)$ is :
   1. A.$48 x+25 y=169$
   2. B.$5 x+16 y=31$
   3. C.$25 x+101 y=176$
   4. D.$4 x+122 y=134$
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   Answer: (A)

   Equation of chord with given middle point $\begin{aligned} & T=S_1 \\ & \Rightarrow \frac{3 x}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \\ & 48 \mathrm{x}+25 \mathrm{y}=144+25 \\ & 48 \mathrm{x}+25 \mathrm{y}=169 \text { Ans. } \end{aligned}$
3. Q3JEE Main 2023 (10 Apr, Shift 1)Auxillary Circle
   Let the ellipse $E:{x}^{2}+9{y}^{2}=9$ intersect the positive $x$- and $y$-axes at the points $A$ and $B$ respectively. Let the major axis of $E$ be a diameter of the circle $C$. Let the line passing through $A$ and $B$ meet the circle $C$ at the point $P$. If the area of the triangle with vertices $A,P$ and the origin $O$ is $\dfrac{m}{n}$, where $m$ and $n$ are coprime, then $m-n$ is equal to
   1. A.$16$
   2. B.$15$
   3. C.$17$
   4. D.$18$
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   Answer: (C)

   Given, Ellipse $E:{x}^{2}+9{y}^{2}=9........\left(i\right)$ Now point $A\left(3,0\right)\&B\left(0,1\right)$ which is the intersection of given ellipse with positive axis, So, equation of line passing through $A\&B$ is given by, $L:\dfrac{x}{3}+\dfrac{y}{1}=1\Rightarrow x=3-3y.......\left(ii\right)$ Now equation of Circle with diametric point $\left(-3,0\right)\&\left(3,0\right)$ is given by, $C:{x}^{2}+{y}^{2}=9........\left(iii\right)$ Let $Q$ be foot of perpendicular from $P$ upon major axis, So, from $\left(ii\right)\&\left(iii\right)$ we get, ${\left(3-3y\right)}^{2}+{y}^{2}=9$ $\Rightarrow y=\dfrac{9}{5},0$ Hence, $PQ=\dfrac{9}{5}$ Now Area of triangle will be, $=\dfrac{1}{2}\times OA\times PQ=\dfrac{1}{2}\times 3\times \dfrac{9}{5}=\dfrac{27}{10}$ Hence, $m-n=17$
4. Q4JEE Main 2022 (28 Jul, Shift 2)Chord of Contact
   Let the tangents at the points $P$ and $Q$ on the ellipse $\dfrac{{x}^{2}}{2}+\dfrac{{y}^{2}}{4}=1$ meet at the point $R\left(\sqrt{2},2\sqrt{2}-2\right)$. If $S$ is the focus of the ellipse on its negative major axis, then $S{P}^{2}+S{Q}^{2}$ is equal to
   Show answer & solution

   Answer: 13

   Given ellipse is $\dfrac{{x}^{2}}{2}+\dfrac{{y}^{2}}{4}=1...........\left(1\right)$, So its eccentricity will be ${a}^{2}={b}^{2}\left(1-{e}^{2}\right)$ $\Rightarrow 2=4\left(1-{e}^{2}\right)$ $\Rightarrow \dfrac{1}{2}=1-{e}^{2}$ $\Rightarrow e=\dfrac{1}{\sqrt{2}}$ So, focus $S$ will be $S\equiv \left(0,-ae\right)\equiv \left(0,-\sqrt{2}\right)$ Now, equation of chord of contact will be $T=0\Rightarrow$ $\dfrac{x}{\sqrt{2}}+\dfrac{\left(2\sqrt{2}-2\right)y}{4}=1$ $\Rightarrow \dfrac{x}{\sqrt{2}}=1-\dfrac{\left(\sqrt{2}-1\right)y}{2}..........\left(2\right)$ Now on solving equation $\left(1\right)\&\left(2\right)$ we get, $\Rightarrow y=0,\sqrt{2}\&x=\sqrt{2},1$ So points $P\&Q$ is given by $P\equiv \left(1,\sqrt{2}\right)\&Q\equiv \left(\sqrt{2},0\right)$ Now using the distance we get, ${\left(SP\right)}^{2}+{\left(SQ\right)}^{2}=13$
5. Q5JEE Main 2021 (31 Aug, Shift 2)Locus
   The locus of mid-points of the line segments joining $\left(-3,-5\right)$ and the points on the ellipse $\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{9}=1$ is :
   1. A.$36{x}^{2}+16{y}^{2}+90x+56y+145=0$
   2. B.$36{x}^{2}+16{y}^{2}+108x+80y+145=0$
   3. C.$9{x}^{2}+4{y}^{2}+18x+8y+145=0$
   4. D.$36{x}^{2}+16{y}^{2}+72x+32y+145=0$
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   Answer: (B)

   Parametric point on the given ellipse is $\left(2\sin \theta ,3\cos \theta \right)$ Let, the mid-point of line segments joining $\left(-3,-5\right)$ and $\left(2\sin \theta ,3\cos \theta \right)$ is $\left(h,k\right)$ Then, by using mid-point formula, we get $\dfrac{2\sin \theta -3}{2}=h,\dfrac{3\cos \theta -5}{2}=k$ $\Rightarrow 2\sin \theta =2h+3,3\cos \theta =2k+5$ $\Rightarrow \sin \theta =\dfrac{2h+3}{2},\cos \theta =\dfrac{2k+5}{3}$ We know, ${\sin }^{2}\theta +{\cos }^{2}\theta =1$ $\Rightarrow {\left(\dfrac{2h+3}{2}\right)}^{2}+{\left(\dfrac{2k+5}{3}\right)}^{2}=1$ $\Rightarrow \dfrac{1}{4}\left[4{h}^{2}+9+12h\right]+\dfrac{1}{9}\left[4{k}^{2}+25+20k\right]=1$ $\Rightarrow 36{h}^{2}+16{k}^{2}+108h+80k+145=0$ Hence, the locus of $\left(x,y\right)$ is $36{x}^{2}+16{y}^{2}+108x+80y+145=0$