Consider a hyperbola H : x2 - 2y2 = 4.

Question

Consider a hyperbola H : x2 - 2y2 = 4. Let the tangent at a
point P(4, $${\sqrt 6 }$$) 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 $$\Delta$$QFR is equal to :

Accepted Answer

Given,

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

$$ \Rightarrow {{{x^2}} \over 4} - {{{y^2}} \over 2} = 1 \Rightarrow {{{x^2}} \over {{{(2)}^2}}} - {{{y^2}} \over {{{(\sqrt 2 )}^2}}} = 1$$

Here, a = 2, $$b = \sqrt 2 $$

$$\therefore$$ $$e = \sqrt {1 + {{{b^2}} \over {{a^2}}}}  = \sqrt {1 + {2 \over 4}}  = \sqrt {1 + {1 \over 2}}  = \sqrt {{3 \over 2}} $$

So, Focus (F) = ($$\pm$$ a e, 0) = ($$\pm$$ $$\sqrt 6 $$, 0)

Now, equation of tangent at $$P(4,\sqrt 6 )$$ is

$x{x_1} - 2y{y_1} = 4$

$$ \Rightarrow x\,.\,4 - 2y\,.\,\sqrt 6  = 4$$

$$ \Rightarrow 4x - 2\sqrt 6 y = 4$$

$$ \Rightarrow 2x - \sqrt 6 y = 2$$ ....... (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 $$x = \sqrt 6 $$

$$\therefore$$ $$R \equiv \left( {\sqrt 6 ,{{2(\sqrt 6  - 1)} \over {\sqrt 6 }}} \right)$$ [putting $$x = \sqrt 6 $$ in Eq. (i), we get $$y = {{2(\sqrt 6  - 1)} \over {\sqrt 6 }}$$]

$$\therefore$$ Area of $$\Delta QFR = {1 \over 2} \times (QF) \times (RF)$$

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

Solutions · 1