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 2 b^{2}=b^{4} $
$\Rightarrow \Delta=b^{4}$
$\left(\because\left(a^{2}, b^{2}\right)\right.$ lies on $\left.\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \Rightarrow a^{2}-b^{2}=1\right)$
Also $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}$
$\Rightarrow e=\sqrt{\frac{2 a^{2}-1}{a^{2}}} $
$\Rightarrow e=\sqrt{2-\frac{1}{a^{2}}}$
$ \Rightarrow 1 < e < \sqrt{2}$
$\Rightarrow e \in(1, \sqrt{2})$