Consider a hyperbola H : x ^{2}-2 y ^{2}=4 . Let the tangent at a point P (4, √{6}) meet x -a

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

hard

Consider a hyperbola $H : x ^{2}-2 y ^{2}=4$. Let the tangent at a point $P (4, \sqrt{6})$ meet the $x$ -axis at $Q$ and latus rectum at $R \left( x _{1}, y _{1}\right), x _{1}>0 .$ If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to ....... .

A

$4 \sqrt{6}$

B

$\sqrt{6}-1$

C

$\frac{7}{\sqrt{6}}-2$

D

$4 \sqrt{6}-1$

(JEE MAIN-2021)

Solution

$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$

$e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{\frac{3}{2}}$

$\therefore$ Focus $F ( ae , 0) \Rightarrow F (\sqrt{6}, 0)$

equation of tangent at $P$ to the hyperbola is

$2 x-y \sqrt{6}=2$

tangent meet $x$ -axis at $Q(1,0)$

And latus rectum $x=\sqrt{6}$ at $R\left(\sqrt{6}, \frac{2}{\sqrt{6}}(\sqrt{6}-1)\right)$

$\therefore$ Area of $\Delta_{ QFR }=\frac{1}{2}(\sqrt{6}-1) \cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)$

$=\frac{7}{\sqrt{6}}-2$

Standard 11

Mathematics

Share

Similar Questions

Consider a hyperbola $\mathrm{H}$ having centre at the origin and foci and the $\mathrm{x}$-axis. Let $\mathrm{C}_1$ be the circle touching the hyperbola $\mathrm{H}$ and having the centre at the origin. Let $\mathrm{C}_2$ be the circle touching the hyperbola $\mathrm{H}$ at its vertex and having the centre at one of its foci. If areas (in sq. units) of $\mathrm{C}_1$ and $\mathrm{C}_2$ are $36 \pi$ and $4 \pi$, respectively, then the length (in units) of latus rectum of $\mathrm{H}$ is

hard

(JEE MAIN-2024)

With one focus of the hyperbola $\frac{{{x^2}}}{9}\,\, – \,\,\frac{{{y^2}}}{{16}}\,\, = \,\,1$ as the centre , a circle is drawn which is tangent to the hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is

normal

For hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\alpha }} – \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1$ which of the following remain constant if $\alpha$ varies

normal

The equation of the tangent to the conic ${x^2} – {y^2} – 8x + 2y + 11 = 0$ at $(2, 1)$ is

medium

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 contacts of the tangents on the hyperbola are

$(A)$ $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ $(B)$ $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

$(C)$ $(3 \sqrt{3},-2 \sqrt{2})$ $(D)$ $(-3 \sqrt{3}, 2 \sqrt{2})$

normal

(IIT-2012)