Hyperbola JEE Main previous year questions with solutions

5 solved JEE Main questions on Hyperbola, free to read — no sign-in needed. The full chapter has 64 questions; sign in to attempt the remaining 59 in the exam simulator.

1. Q1JEE Main 2026 (02 Apr, Shift 2)Chord with given Middle Point
   Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$. Then the area of the triangle OPQ equals:
   1. A.$\dfrac{3}{2}$
   2. B.$\dfrac{5}{2}$
   3. C.$\dfrac{7}{2}$
   4. D.$\dfrac{9}{2}$
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   Answer: (C)

   The equation of the chord of the hyperbola $xy = 12$ with midpoint $(x_1, y_1)$ is given by $T = S_1$. $\dfrac{x y_1 + y x_1}{2} - 12 = x_1 y_1 - 12$ $x y_1 + y x_1 = 2 x_1 y_1$ Substituting the midpoint $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$: $x\left(-\dfrac{1}{2}\right) + y\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)$ $-x + y = -1 \Rightarrow y = x - 1$ To find the coordinates of P and Q, substitute $y = x - 1$ into the equation of the hyperbola $xy = 12$: $x(x - 1) = 12$ $x^2 - x - 12 = 0$ $(x - 4)(x + 3) = 0 \Rightarrow x = 4, -3$ For $x = 4$, $y = 3$. For $x = -3$, $y = -4$. Thus, the coordinates of P and Q are $(4, 3)$ and $(-3, -4)$. The area of triangle OPQ with vertices $(0,0)$, $(4,3)$, and $(-3,-4)$ is: $\text{Area} = \dfrac{1}{2} |x_P y_Q - x_Q y_P|$ $\text{Area} = \dfrac{1}{2} |(4)(-4) - (-3)(3)|$ $\text{Area} = \dfrac{1}{2} |-16 + 9| = \dfrac{7}{2}$ Answer: $\dfrac{7}{2}$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Equation of hyperbola
   <p>Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be $(-5,0)$ and $5 x+9=0$, respectively. If the product of the focal distances of a point $(\alpha, 2 \sqrt{5})$ on the hyperbola is $p$, then $4 p$ is equal to</p>
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   Answer: 189

   <p>Equation of hyperbola is $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ Directrix: $\mathrm{x}=\frac{-9}{5}$ and corresponding foci $(-5,0)$ $\Rightarrow-\frac{\mathrm{a}}{\mathrm{e}}=-\frac{9}{5}$ and $-\mathrm{ae}=-5$ $\Rightarrow \frac{9 e^2}{5}=5 \Rightarrow e=\sqrt{\frac{25}{9}}=\frac{5}{3} \Rightarrow a=3$ $\therefore \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=9\left(\frac{25}{9}-1\right)=16$ Hyperbola $\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1$ $(\alpha, 2 \sqrt{5})$ lie on it $\Rightarrow \frac{\alpha^2}{9}-\frac{20}{16}=1 \Rightarrow \alpha^2=\frac{36}{16} \times 9=\frac{81}{4}$ Product for distance of $\left(\mathrm{x}_1 \mathrm{y}_1\right)$ from the two foci $\begin{aligned} &amp; =\left(e x_1+a\right)\left|e x_1-a\right| \\ &amp; =e^2 x_1^2-a^2 \end{aligned}$ For $(\alpha, 2 \sqrt{5}) \Rightarrow \mathrm{P}=\frac{25}{9} \cdot \frac{81}{4}-9=\frac{189}{4}$ $4 \mathrm{P}=189$</p>
3. Q3JEE Main 2023 (06 Apr, Shift 2)Mixed Questions of Ellipse and Hyperbola
   Let the eccentricity of an ellipse $\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$ is reciprocal to that of the hyperbola $2{x}^{2}-2{y}^{2}=1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is _____.
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   Answer: 2

   Given, The eccentricity of an ellipse $\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$ is reciprocal to that of the hyperbola $2{x}^{2}-2{y}^{2}=1$, Now eccentricity of rectangular hyperbola $2{x}^{2}-2{y}^{2}=1$ is ${e}_{H}=\sqrt{2}$ So, ${e}_{e}=\dfrac{1}{\sqrt{2}}$ Now, focus of hyperbola $=\left(\pm 1,0\right)$ Now given that both curve intersect orthogonally, so ellipse and hyperbola are confocal So, for ellipse $a{e}_{e}=1\Rightarrow a=\sqrt{2}$ Now length of latusrectum $L.R.=\dfrac{2{b}^{2}}{a}=2a\left(1-{{e}_{e}}^{2}\right)$ $=2\sqrt{2}.\dfrac{1}{2}=\sqrt{2}$
4. Q4JEE Main 2021 (25 Feb, Shift 1)Locus
   The locus of the point of intersection of the lines $\left(\sqrt{3}\right)kx+ky-4\sqrt{3}=0$ and $\sqrt{3}x-y-4\left(\sqrt{3}\right)k=0$ is a conic, whose eccentricity is
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   Answer: 2

   The point of intersection of the lines $\left(\sqrt{3}\right)kx+ky-4\sqrt{3}=0$ and $\sqrt{3}x-y-4\left(\sqrt{3}\right)k=0$. $k=\dfrac{4\sqrt{3}}{\sqrt{3}x+y}=\dfrac{\sqrt{3}x-y}{4\sqrt{3}}$ $\Rightarrow 3{x}^{2}-{y}^{2}=48$ $\Rightarrow \dfrac{{x}^{2}}{16}-\dfrac{{y}^{2}}{48}=1$ Which is a hyperbola. ${b}^{2}={a}^{2}\left({e}^{2}-1\right)$ Now, $48=16\left({e}^{2}-1\right)$ $\Rightarrow e=\sqrt{4}=2$
5. Q5JEE Main 2026 (06 Apr, Shift 1)Equation of hyperbola
   If the eccentricity $e$ of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, passing through $(6, 4\sqrt{3})$, satisfies $15(e^2 + 1) = 34e$, then the length of the latus rectum of the hyperbola $\dfrac{x^2}{b^2} - \dfrac{y^2}{2(a^2+1)} = 1$ is:
   1. A.$10$
   2. B.$20$
   3. C.$25$
   4. D.$30$
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   Answer: (A)

   The given equation of the hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$. Since it passes through $(6, 4\sqrt{3})$, we have: $\dfrac{36}{a^2} - \dfrac{48}{b^2} = 1$ The eccentricity $e$ satisfies $15(e^2 + 1) = 34e$. $15e^2 - 34e + 15 = 0$ $(3e - 5)(5e - 3) = 0$ Since the eccentricity of a hyperbola is $e \gt 1$, we get $e = \dfrac{5}{3}$. We know that $b^2 = a^2(e^2 - 1)$. $b^2 = a^2 \left(\dfrac{25}{9} - 1\right) = \dfrac{16}{9}a^2$ Substituting $b^2$ into the first equation: $\dfrac{36}{a^2} - \dfrac{48}{\dfrac{16}{9}a^2} = 1$ $\dfrac{36}{a^2} - \dfrac{27}{a^2} = 1$ $\dfrac{9}{a^2} = 1 \Rightarrow a^2 = 9$ Then, $b^2 = \dfrac{16}{9} \times 9 = 16$. The second hyperbola is given by $\dfrac{x^2}{b^2} - \dfrac{y^2}{2(a^2+1)} = 1$. Substituting the values of $a^2$ and $b^2$: $\dfrac{x^2}{16} - \dfrac{y^2}{2(9+1)} = 1$ $\dfrac{x^2}{16} - \dfrac{y^2}{20} = 1$ Here, $A^2 = 16 \Rightarrow A = 4$ and $B^2 = 20$. The length of the latus rectum is $\dfrac{2B^2}{A}$. Length of latus rectum $= \dfrac{2 \times 20}{4} = 10$. Answer: $10$