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.

Answers checked against the official answer keys.

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.$\dfrac{9}{2}$
   Show answer & solution

   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>
   Show answer & solution

   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 2022 (25 Jul, Shift 2)Mixed Questions of Ellipse and Hyperbola
   Let the foci of the ellipse $\dfrac{{x}^{2}}{16}+\dfrac{{y}^{2}}{7}=1$ and the hyperbola $\dfrac{{x}^{2}}{144}-\dfrac{{y}^{2}}{\alpha }=\dfrac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:
   1. A.$\dfrac{32}{9}$
   2. B.$\dfrac{18}{5}$
   3. C.$\dfrac{27}{4}$
   4. D.$\dfrac{27}{10}$
   Show answer & solution

   Answer: (D)

   Given equation of ellipse $\dfrac{{x}^{2}}{16}+\dfrac{{y}^{2}}{7}=1$ Now finding eccentricity $=\sqrt{1-\dfrac{7}{16}}=\dfrac{3}{4}$ So, foci $\equiv \left(\pm ae,0\right)\equiv \left(\pm 3,0\right)$ Now, hyperbola: $\dfrac{{x}^{2}}{\left(\dfrac{144}{25}\right)}-\dfrac{{y}^{2}}{\left(\dfrac{\alpha }{25}\right)}=1$ Eccentricity will be $=\sqrt{1+\dfrac{\alpha }{144}}=\dfrac{1}{12}\sqrt{144+\alpha }$ Foci $\equiv \left(\pm ae,0\right)\equiv \left(\pm \dfrac{12}{5}\cdot \dfrac{1}{12}\sqrt{144+\alpha },0\right)$ Given foci coincide then $3=\dfrac{1}{5}\sqrt{144+\alpha }\Rightarrow \alpha =81$ Hence, hyperbola is $\dfrac{{x}^{2}}{{\left(\dfrac{12}{5}\right)}^{2}}-\dfrac{{y}^{2}}{{\left(\dfrac{9}{5}\right)}^{2}}=1$ Length of latus rectum $=2\dfrac{{b}^{2}}{a}=2\cdot \dfrac{\dfrac{81}{25}}{\dfrac{12}{5}}=\dfrac{27}{10}$
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
   Show answer & solution

   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.$30$
   Show answer & solution

   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: $10$