Consider a hyperbola H: x²-2 y²=4. Let the tangent at a point P(4, {6}) meet the x-axis at Q and latus rectum at R(x₁, y₁), x₁>0. If F is a focus of H which is nearer to the point P, then the area of QFR is equal to [18 Mar 2021 Shift 2]

Correct Answer is : 76−2\; {7}{{6}} - 267−2

Correct Answer is : 76−2\; {7}{{6}} - 267−2

Test Index

Hyperbola

Section: Mathematics

Question : 63 of 98

Marks: +1, -0

Consider a hyperbola

H: x^{2}-2 y^{2}=4

P(4, \sqrt{6})

x

Q

R(x_{1}, y_{1}), x_{1}>0

F

H

P

\triangle QFR

Solution:

Given,

x^{2}-2 y^{2}=4

\Rightarrow\;\;\; \frac{x^{2}}{4} - \; \frac{y^{2}}{2} = 1 \Rightarrow \; \frac{x^{2}}{(2)^{2}} - \; \frac{y^{2}}{(\sqrt{2})^{2}} = 1

Here,

a=2, b=\sqrt{2}

\therefore \;\;\; e = \sqrt{1 + \; \frac{b^{2}}{a^{2}}} = \sqrt{1 + \; \frac{2}{4}} = \sqrt{1 + \; \frac{1}{2}} = \sqrt{\; \frac{3}{2}}

So, Focus

(F)=(\pm a e, 0)=(\pm \sqrt{6}, 0)

Now, equation of tangent at

P(4, \sqrt{6})

4 x - 2 \sqrt{6} y = 4 \;\; [ \because x \cdot x - 2 \cdot y \cdot y = 4, x \cdot 4 - 2 \cdot y \cdot \sqrt{6} = 4]

\Rightarrow \;\; 2 x - \sqrt{6 y} = 2

...(i)

Putting

y=0

in Eq. (i), we get

x

-intercept of tangent i.e.

x=1

\therefore \;\; Q \equiv (1,0)

Hence, equation of corresponding latus rectum is

x=\sqrt{6}

\therefore R \equiv \left(\sqrt{6}, \; \frac{2(\sqrt{6}-1)}{\sqrt{6}}\right)

[putting

x=\sqrt{6}

in Eq. (i), we get

y = \; \frac{2(\sqrt{6}-1)}{\sqrt{6}}]

\therefore

Area of

\triangle QFR = \; \frac{1}{2} \times (QF) \times (RF)

= \; \frac{1}{2} (\sqrt{6}-1) \times \; \frac{2(\sqrt{6}-1)}{\sqrt{6}} = \; \frac{(\sqrt{6}-1)^{2}}{\sqrt{6}} = \left( \; \frac{7}{\sqrt{6}} - 2\right)

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