Question

Let $a$ and $b$ be positive real numbers such that $a>1$ and $b<a.$ Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ Suppose the tangent to the hyperbola at $P$ passes through the point $(1,0),$ and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P,$ the normal at $P$ and the $x-$axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?

• A. $1<e<\sqrt{2}$
• B. $\sqrt{2}<e<2$
• C. $\Delta =a^4$
• D. $\Delta =b^4$

Answer 1

Answer: (A), (D)

$\Delta =b^4$

Normal cuts equal intercepts.
$\therefore M_N=-1$
and $M_T=1$

Equation of tangent at $P$ is
$\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$
The tangent passes through $(1,0)$
$\Rightarrow \sec \theta =a$
$M_T=1$
$\Rightarrow \frac{b\sec \theta }{a\tan \theta }=1$
$\Rightarrow b=\tan \theta [\because a=\sec \theta ]$

$b^2=a^2(e^2-1)$
$\Rightarrow e^2-1={\sin }^2\theta$
$\Rightarrow e^2=1+{\sin }^2\theta$
Since $0<\theta <\frac{\pi }{2},$
$1<e^2<2$
$\Rightarrow 1<e<\sqrt{2}$

Area, $\Delta =\frac{1}{2}\left(AP\right)\left(AP\right)[\because AP=BP]$
$=\frac{1}{2}[{(1-{\sec }^2\theta )}^2+{\left({\tan }^2\theta \right)}^2]$
$={\tan }^4\theta =b^4$