Consider a hyperbola H:x^2-2y^2=4.

Question

Consider a hyperbola $H:x^2-2y^2=4$. Let the tangent at a point $P(4,\sqrt6)$ meet the $x$-axis at $Q$ and the latus rectum at $R(x_1,y_1)$, $x_1>0$. If $F$ is the focus of $H$ nearer to $P$, then the area of $	riangle QFR$ is equal to

Accepted Answer

$\dfrac{x^2}4-\dfrac{y^2}2=1$, $c^2=6$, foci $(\pm\sqrt6,0)$; nearer focus to $P(4,\sqrt6)$ is $F=(\sqrt6,0)$.

Tangent at $P$: $\dfrac{4x}4-\dfrac{\sqrt6y}2=1\Rightarrow x-\dfrac{\sqrt6}2y=1$. At $y=0$: $Q=(1,0)$.

At $x=\sqrt6$ (latus rectum): $\sqrt6-\dfrac{\sqrt6}2y=1\Rightarrow y=\dfrac{2(\sqrt6-1)}{\sqrt6}=2-\dfrac{\sqrt6}3=\dfrac{6-\sqrt6}3=y_1$.

$$\text{Area}=\frac12(FQ)(y_1)=\frac12(\sqrt6-1)\cdot\frac{6-\sqrt6}3=\frac{7\sqrt6-12}6=\frac7{\sqrt6}-2.$$

Solutions · 1