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)

As in first quadrant if normal at $P$ is making equal intercepts on axes,
then slope of the normal $=-1$
$\Rightarrow$ slope of tangent $=1$
$\Rightarrow$ equation of tangent at $P:\frac{y-0}{x-1}=1$
and equation of tangent at $\left(x_1,y_1\right):\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$
$\Rightarrow x_1=a^2$ and $y_1=b^2$
$\Rightarrow P\left(x_1,y_1\right)\equiv P\left(a^2,b^2\right)$
$\Rightarrow$ equation of normal at $P:x+y=a^2+b^2$
$\Rightarrow B\equiv \left(a^2+b^2,0\right)$
$\Rightarrow \Delta =\frac{1}{2}\times b^2\times \left(a^2+b^2-1\right)$
$=\frac{1}{2}\times b^2\times 2b^2=b^4$
$\Rightarrow \Delta =b^4$
$(\because \left(a^2,b^2\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow a^2-b^2=1)$
Also $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{a^2+b^2}{a^2}}$
$\Rightarrow e=\sqrt{\frac{2a^2-1}{a^2}}$
$\Rightarrow e=\sqrt{2-\frac{1}{a^2}}$
$\Rightarrow 1<e<\sqrt{2}$
$\Rightarrow e\in (1,\sqrt{2})$