Hyperbola JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Hyperbola, free to read — no sign-in needed. The full chapter has 13 questions; sign in to attempt the remaining 9 in the exam simulator.

1. Q1JEE Advanced Adv 2012 (Paper 1)
   Tangents are drawn to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$, parallel to the straight line $2 x-y=1$. The points of contact of the tangents on the hyperbola are
   1. A.$\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
   2. B.$\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
   3. C.$(3 \sqrt{3},-2 \sqrt{2})$
   4. D.$(-3 \sqrt{3}, 2 \sqrt{2})$
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   Answer: A,B

   If slope of tangents to hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $m$, then equations of tangent to the hyperbola is $y=m x \pm \sqrt{a^{2} m^{2}-b^{2}} \quad$ with the points of contact $\quad\left(\quad \pm a^{2} m\right.$ $\left.\frac{\pm \sqrt{a^{2} m^{2}-b^{2}}}{\sqrt{a^{2} m^{2}-b^{2}}}\right)$ $\therefore$ Tangent to hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ is parallel to $2 x-y=1$, $\therefore$ Slope of tangent $=2$ $\therefore$ Points of contact are $\left(\frac{\pm 9 \times 2}{\sqrt{9 \times 4-4}}, \frac{\pm 4}{\sqrt{9 \times 4-4}}\right)$ i.e. $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{-9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)$
2. Q2JEE Advanced Adv 2011 (Paper 2)
   Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $X$-axis at $(9,0)$, then the eccentricity of the hyperbola is
   1. A.$\sqrt{\frac{5}{2}}$
   2. B.$\sqrt{\frac{3}{2}}$
   3. C.$\sqrt{2}$
   4. D.$\sqrt{3}$
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   Answer: (B)

   Equation of normal to hyperbola at $\left(x_1, y_1\right)$ is $$\begin{gathered} \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \\ \therefore \text { At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2 \end{gathered}$$ It passes throught $(9,0)$. $$\begin{aligned} & \text { Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b^2 \\ & \Rightarrow \quad \frac{3 a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2 \\ & \therefore \quad e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}} \end{aligned}$$
3. Q3JEE Advanced Adv 2011 (Paper 1)
   Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellipse $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then
   1. A.the equation of the hyperbola is $\frac{x^2}{3^2}-\frac{y^2}{2^2}=1$
   2. B.a focus of the hyperbola is $(2,0)$
   3. C.the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
   4. D.the equation of the hyperbola is $x^2-3 y^2=3$
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   Answer: B,D

   Here, equation of ellipse $$\begin{aligned} & \frac{x^2}{4}+\frac{y^2}{1}=1 \\ & \Rightarrow \quad e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4} \\ & \therefore \quad e=\frac{\sqrt{3}}{2} \text { and focus }(\pm a e, 0) \\ & =(\pm \sqrt{3}, 0) \\ & \end{aligned}$$ For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $e_1^2=1+\frac{b^2}{a^2}$ where, $e_1^2=\frac{1}{e^2}=\frac{4}{3}$ $$\Rightarrow \quad 1+\frac{b^2}{a^2}=\frac{4}{3} \Rightarrow \frac{b^2}{a^2}=\frac{1}{3}$$ and hyperbola passes through $(\pm \sqrt{3}, 0)$. Now, $\quad \frac{3}{a^2}=1 \Rightarrow a^2=3$ From Eqs. (i) and (ii), we get $b^2=1$ $\therefore$ Equation of hyperbola is $$\frac{x^2}{3}-\frac{y^2}{1}=1$$ Focus is $(\pm a e, 0)$. Now, $\quad\left(\pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0\right) \Rightarrow(\pm 2,0)$ Hence, both options (b) and (d) are correct.
4. Q4JEE Advanced Adv 2010 (Paper 1)
   Paragraph: The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question: Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
   1. A.$2 x-\sqrt{5} y-20=0$
   2. B.$2 x-\sqrt{5} y+4=0$
   3. C.$3 x-4 y+8=0$
   4. D.$4 x-3 y+4=0$
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   Answer: (B)

   Equation of tangent to hyperbola having slope $m$ is $$y=m x+\sqrt{9 m^2-4}$$ Equation of tangent to circle is $$y=m(x-4)+\sqrt{16 m^2+16}$$ Eqs. (i) and (ii) will be identical for $m=\frac{2}{\sqrt{5}}$ satisfy. $\therefore$ Equation of common tangent is $2 x-\sqrt{5} y+4=0$