Ellipse JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Ellipse, free to read — no sign-in needed. The full chapter has 22 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Passage: Consider the ellipses given by $x^2 + 4y^2 = 1$ and $4x^2 + y^2 = 1$. Question: Let $P$ be the point in the first quadrant where the given ellipses intersect. If $\theta$ is the acute angle between the tangents to the given ellipses at the point $P$, then the value of $4\tan\theta$ is ___________.
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   Answer: 7.5

   The given equations of the ellipses are $x^2 + 4y^2 = 1$ and $4x^2 + y^2 = 1$. To find the point of intersection, we equate the two expressions: $x^2 + 4y^2 = 4x^2 + y^2$ $3x^2 = 3y^2 \Rightarrow x = y$ (since $P$ lies in the first quadrant) Substituting $x = y$ into the first ellipse equation: $x^2 + 4x^2 = 1 \Rightarrow 5x^2 = 1 \Rightarrow x = \dfrac{1}{\sqrt{5}}$ Thus, the point of intersection is $P\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}\right)$. Differentiating $x^2 + 4y^2 = 1$ with respect to $x$ gives the slope of the tangent to the first ellipse: $2x + 8y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{4y}$ At $P\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}\right)$, the slope is $m_1 = -\dfrac{1}{4}$. Differentiating $4x^2 + y^2 = 1$ with respect to $x$ gives the slope of the tangent to the second ellipse: $8x + 2y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{4x}{y}$ At $P\left(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}\right)$, the slope is $m_2 = -4$. The acute angle $\theta$ between the tangents is given by: $\tan\theta = \left| \dfrac{m_1 - m_2}{1 + m_1 m_2} \right|$ $\tan\theta = \left| \dfrac{-\dfrac{1}{4} - (-4)}{1 + \left(-\dfrac{1}{4}\right)(-4)} \right| = \left| \dfrac{\dfrac{15}{4}}{2} \right| = \dfrac{15}{8}$ Therefore, the value of $4\tan\theta$ is: $4\tan\theta = 4 \times \dfrac{15}{8} = \dfrac{15}{2} = 7.5$ Answer: $7.5$
2. Q2JEE Advanced Adv 2026 (Paper 2)
   Consider the ellipse $E$ given by $\dfrac{x^2}{18} + \dfrac{y^2}{12} = 1$. Let $H$ be the hyperbola whose eccentricity is the reciprocal of the eccentricity of $E$ and whose foci are the same as that of $E$. Let $P$ and $Q$ be the points of intersection of $H$ and the parabola $\sqrt{5}\, y = x^2$ in the first quadrant. Let $d$ be the distance between $P$ and $Q$. If $a$ and $b$ are the integers such that $d^2 = a + b\sqrt{5}$, then the value of $a - b$ is __________.
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   Answer: 18

   For the ellipse $E: \dfrac{x^2}{18} + \dfrac{y^2}{12} = 1$, we have $a_E^2 = 18$ and $b_E^2 = 12$. The eccentricity $e_E$ is given by: $e_E = \sqrt{1 - \dfrac{b_E^2}{a_E^2}} = \sqrt{1 - \dfrac{12}{18}} = \dfrac{1}{\sqrt{3}}$ The foci of the ellipse $E$ are $(\pm a_E e_E, 0) = \left(\pm \sqrt{18} \times \dfrac{1}{\sqrt{3}}, 0\right) = (\pm \sqrt{6}, 0)$. For the hyperbola $H$, the eccentricity $e_H$ is the reciprocal of $e_E$, so $e_H = \sqrt{3}$. Since $H$ has the same foci as $E$, its foci are $(\pm \sqrt{6}, 0)$. Let the equation of $H$ be $\dfrac{x^2}{A^2} - \dfrac{y^2}{B^2} = 1$. The foci are $(\pm A e_H, 0)$, which gives $A \sqrt{3} = \sqrt{6} \Rightarrow A = \sqrt{2} \Rightarrow A^2 = 2$. Also, $B^2 = A^2(e_H^2 - 1) = 2(3 - 1) = 4$. Thus, the equation of the hyperbola $H$ is $\dfrac{x^2}{2} - \dfrac{y^2}{4} = 1$. To find the points of intersection of $H$ and the parabola $\sqrt{5}y = x^2$, we substitute $x^2 = \sqrt{5}y$ into the equation of $H$: $\dfrac{\sqrt{5}y}{2} - \dfrac{y^2}{4} = 1$ Multiplying by $4$, we get: $2\sqrt{5}y - y^2 = 4 \Rightarrow y^2 - 2\sqrt{5}y + 4 = 0$ Solving for $y$ using the quadratic formula: $y = \dfrac{2\sqrt{5} \pm \sqrt{20 - 16}}{2} = \sqrt{5} \pm 1$ Let $y_1 = \sqrt{5} - 1$ and $y_2 = \sqrt{5} + 1$. Both are positive. The corresponding $x^2$ values are: $x_1^2 = \sqrt{5}(\sqrt{5} - 1) = 5 - \sqrt{5}$ $x_2^2 = \sqrt{5}(\sqrt{5} + 1) = 5 + \sqrt{5}$ Since the points $P$ and $Q$ lie in the first quadrant, $x \gt 0$. Thus, $x_1 = \sqrt{5 - \sqrt{5}}$ and $x_2 = \sqrt{5 + \sqrt{5}}$. The square of the distance $d$ between $P$ and $Q$ is: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ First, we evaluate $(x_2 - x_1)^2$: $(x_2 - x_1)^2 = x_2^2 + x_1^2 - 2x_1x_2$ $x_2^2 + x_1^2 = (5 + \sqrt{5}) + (5 - \sqrt{5}) = 10$ $x_1x_2 = \sqrt{(5 - \sqrt{5})(5 + \sqrt{5})} = \sqrt{25 - 5} = \sqrt{20} = 2\sqrt{5}$ So, $(x_2 - x_1)^2 = 10 - 4\sqrt{5}$. Next, we evaluate $(y_2 - y_1)^2$: $(y_2 - y_1)^2 = ((\sqrt{5} + 1) - (\sqrt{5} - 1))^2 = 2^2 = 4$ Adding these together gives $d^2$: $d^2 = 10 - 4\sqrt{5} + 4 = 14 - 4\sqrt{5}$ Comparing this with $d^2 = a + b\sqrt{5}$, we get $a = 14$ and $b = -4$. Therefore, $a - b = 14 - (-4) = 18$. Answer: $18$
3. Q3JEE Advanced Adv 2015 (Paper 2)
   Let ${E}_{1}$ and ${E}_{2}$ be two ellipse whose centers are at the origin. The major axes of ${E}_{1}and{E}_{2}$ lie along the x - axis and the y - axis, respectively. Let S be the circle ${x}^{2}+{\left(y-1\right)}^{2}=2.$ The straight line $x+y=3$ touches the curves $S,{E}_{1}$ and ${E}_{2}$ at $P,Q$ and $R,$ respectively. Suppose that $PQ=PR=\dfrac{2\sqrt{2}}{3}.$ If${e}_{1}$ and ${e}_{2}$ are the eccentricities of ${E}_{1}$ and ${E}_{2},$respectively, then the correct expression(s) is(are)
   1. A.${e}_{1}^{2}+{e}_{2}^{2}=\dfrac{43}{40}$
   2. B.${e}_{1}{e}_{2}=\dfrac{\sqrt{7}}{2\sqrt{10}}$
   3. C.$\left|{e}_{1}^{2}-{e}_{2}^{2}\right|=\dfrac{5}{8}$
   4. D.${e}_{1}{e}_{2}=\dfrac{\sqrt{3}}{4}$
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   Answer: A,B

   Let Q be $({x}_{1},{y}_{1})$ So equation tangent at Q will be $x{x}_{1}+y{y}_{1}-\left(y+{y}_{1}\right)-1=0$ Comparing with $x+y=3$ $\left({x}_{1}{y}_{1}\right)\equiv (1,2)$ So R and Q will be $\left(1\pm \dfrac{2\sqrt{2}}{3}\cos ⁡\dfrac{3\pi }{4},2\pm \dfrac{2\sqrt{2}}{3}\sin ⁡\dfrac{3\pi }{4}\right)$ $\Rightarrow Q\left(\dfrac{5}{3},\dfrac{4}{3}\right)and\left(\dfrac{1}{3},\dfrac{8}{3}\right)$ Let Q lies on $\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$ So tangent at P is $\dfrac{5x}{3{a}^{2}}+\dfrac{4y}{3{b}^{2}}=1$ Comparing with $x+y=3$ ${a}^{2}=5,{b}^{2}=4\Rightarrow {e}_{1}=\dfrac{1}{\sqrt{5}}$ And R lies on $\dfrac{{x}^{2}}{{a}_{1}^{2}}+\dfrac{{y}^{2}}{{b}_{1}^{2}}=1$ So tangent at $\dfrac{x}{{3a}_{1}^{2}}+\dfrac{8y}{{3b}_{1}^{2}}=3$ Comparing with $x+y=3$ $\Rightarrow {a}_{1}^{2}=1,{b}_{1}^{2}=8\Rightarrow {e}_{2}=\sqrt{\dfrac{7}{8}}$ $\Rightarrow {e}_{1}^{2}+{e}_{2}^{2}=\dfrac{43}{40}and{e}_{1}{e}_{2}=\dfrac{\sqrt{7}}{2\sqrt{10}}$
4. Q4JEE Advanced Adv 2010 (Paper 2)
   Paragraph: Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question: The equation of the locus of the point whose distance from the point $P$ and the line $A B$ are equal, is
   1. A.$9 x^2+y^2-6 x y-54 x-62 y$ $$+241=0$$
   2. B.$x^2+9 y^2+6 x y-54 x+62 y$ $$-241=0$$
   3. C.$9 x^2+9 y^2-6 x y-54 x-62 y$ $$-241=0$$
   4. D.$x^2+y^2-2 x y+27 x+31 y$ $$-120=0$$
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   Answer: (A)

   Equation of $A B$ is $y-0=-\frac{1}{3}(x-3)$ $$\begin{gathered} x+3 y-3=0 \\ |x+3 y-3|^2=10\left[(x-3)^2+(y-4)^2\right] \end{gathered}$$ (Look at coefficient of $x^2$ and $y^2$ in the answers)
5. Q5JEE Advanced Adv 2009 (Paper 2)
   An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then
   1. A.equation of ellipse is $x^2+2 y^2=2$
   2. B.the foci of ellipse are $(\pm 1,0)$
   3. C.equation of ellipse is $x^2+2 y^2=4$
   4. D.the foci of ellipse are $(\pm \sqrt{2}, 0)$
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   Answer: A,B

   Given, $2 x^2-2 y^2=1$ $$\Rightarrow \quad \frac{x^2}{\left(\frac{1}{2}\right)}-\frac{y^2}{\left(\frac{1}{2}\right)}=1$$ Eccentricity of hyperbola $=\sqrt{2}$ So, eccentricity of ellipse $=1 / \sqrt{2}$ Let equation of ellipse be $$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \\ & \Rightarrow \quad \frac{b^2}{a^2}=\frac{1}{2} \Rightarrow a^2=2 b^2 \\ & \therefore \quad x^2+2 y^2=2 b^2 \\ & \end{aligned}$$ Let ellipse and hyperbola intersect at $$A\left(\frac{1}{\sqrt{2}} \sec \theta, \frac{1}{\sqrt{2}} \tan \theta\right)$$ On differentiating Eq. (i), $$4 x-4 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x}{y}$$ $$\therefore \quad\left(\frac{d y}{d x}\right)_{\text {at } A}=\frac{\sec \theta}{\tan \theta}=\operatorname{cosec} \theta$$ On differentiating Eq. (ii), $$2 x+4 y \frac{d y}{d x}=0$$ $$\therefore \quad\left(\frac{d y}{d x}\right)_{\text {at } A}=-\frac{x}{2 y}=-\frac{1}{2} \operatorname{cosec} \theta$$ Since, ellipse and hyperbola are orthogonal. $$\begin{aligned} & \therefore \quad-\frac{1}{2} \operatorname{cosec}^2 \theta=-1 \\ & \Rightarrow \quad \operatorname{cosec}^2 \theta=2 \Rightarrow \theta=\pm \frac{\pi}{4} \\ & \therefore A\left(1, \frac{1}{\sqrt{2}}\right) \text { or }\left(1,-\frac{1}{\sqrt{2}}\right) \end{aligned}$$ From Eq. (i), $1+2\left(\frac{1}{\sqrt{2}}\right)^2=2 b^2$ $\Rightarrow \quad b^2=1$ Equation of ellipse is $x^2+2 y^2=2$ Coordinate of foci $(\pm a e, 0)$ $$=\left(\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0\right)=(\pm 1,0)$$ Hence, options (a) and (b) are correct. If major axis is along $Y$-axis, then $$\begin{aligned} & \frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}} \Rightarrow b^2=2 a^2 \\ & \therefore 2 x^2+y^2=2 a^2 \Rightarrow y^{\prime}=-\frac{2 x}{y} \\ & \Rightarrow \quad y^{\prime}\left(\frac{1}{\sqrt{2}} \sec \theta \frac{1}{\sqrt{2}} \tan \theta\right)=\frac{-2}{\sin \theta} \\ & \end{aligned}$$ As ellipse and hyperbola are orthogonal. $$\begin{array}{ll} \therefore & -\frac{2}{\sin \theta} \cdot \operatorname{cosec} \theta=-1 \\ \Rightarrow & \operatorname{cosec}^2 \theta=1 \Rightarrow \theta=\pm \frac{\pi}{4} \\ \therefore & 2 x^2+y^2=2 a^2 \Rightarrow 2+\frac{1}{2}=2 a^2 \\ \Rightarrow & a^2=\frac{5}{4} \Rightarrow 2 x^2+y^2=\frac{5}{2} \end{array}$$ Corresponding foci are $(0, \pm 1)$.