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 $\left(1,0\right)$, and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $∆$ 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. $∆=a^4$
• D. $∆=b^4$

Answer 1

Answer: (A), (D)

$∆=b^4$

Explanation for the correct option.

Given: The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

Now tangent for the hyperbola at $\left(x_1,y_1\right)$ is given as $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$.

Let $P\left(x_1,y_1\right)=P\left(asec\theta ,b\tan \theta \right)$ be the point.

$\begin{array}{rcl}\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}& =& 1\\ & \Rightarrow & \frac{axsec\theta }{a^2}-\frac{by\tan \theta }{b^2}=1\\ & \Rightarrow & \frac{xsec\theta }{a}-\frac{y\tan \theta }{b}=1\end{array}$

Since the point $\left(1,0\right)$ satisfies the tangent equation.

$\begin{array}{rcl}\frac{sec\theta }{a}-0& =& 1\\ & \Rightarrow & \frac{sec\theta }{a}=1\\ & \Rightarrow & sec\theta =a\end{array}$

Now the slope of normal is given as

$\begin{array}{rcl}\text{Slope of normal}& =& -\frac{\text{y-intersept}}{\text{x-intersept}}\\ & =& -1(\mathrm{As}\mathrm{the}\mathrm{intercepts}\mathrm{are}\mathrm{equal})\end{array}$

Also

$\begin{array}{rcl}\text{Slope of tangent}& =& \frac{-1}{\mathrm{Slope}\mathrm{of}\text{normal}}\\ & =& \frac{-1}{-1}\\ & =& 1\end{array}$

Also, we have slope of tangent as

$\begin{array}{rcl}\frac{b\tan \theta -0}{se{c}^2\theta -1}& =& 1\\ & \Rightarrow & \frac{b\tan \theta }{{\tan }^2\theta }=1\\ & \Rightarrow & \frac{b}{\tan \theta }=1\\ & \Rightarrow & b=\tan \theta \end{array}$

So, the value of $a$ is $sec\theta$ and $b$ is $\tan \theta$, then point $P$ is $\left(se{c}^2\theta ,{\tan }^2\theta \right)$.

The eccentricity of the hyperbola is given as,

$b^2=a^2\left(e^2-1\right)$

Substitute $a$ as $sec\theta$ and $b$ as $\tan \theta$ in the above eccentricity.

$$\begin{gathered} {\tan }^2\theta =se{c}^2\theta \left(e^2-1\right) \\ \Rightarrow \frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{1}{{\cos }^2\theta }\left(e^2-1\right) \\ \Rightarrow \left(e^2-1\right)={\sin }^2\theta \end{gathered}$$

We know that, $0\le {\sin }^2\le 1$, then we have $1<e<\sqrt{2}$.

It is also given that $∆$ is the area of the triangle formed by the tangent at $P$, the normal at $P$ and the x-axis.

$\begin{array}{rcl}∆& =& areaofAPQ\\ & =& \frac{1}{2}\times \alpha \times \alpha \\ & =& \frac{1}{2}{\alpha }^2\end{array}$

Now find the value of $\alpha$.

$\begin{array}{rcl}\alpha & =& \sqrt{{\left({\tan }^2\theta -0\right)}^2+{\left(se{c}^2\theta -1\right)}^2}\\ & =& \sqrt{{\left({\tan }^2\theta \right)}^2+{\left({\tan }^2\theta \right)}^2}\\ & =& \sqrt{2{\left({\tan }^2\theta \right)}^2}\end{array}$

Take square on both sides of the above equation.

$\begin{array}{rcl}{\alpha }^2& =& {\left(\sqrt{2{\left({\tan }^2\theta \right)}^2}\right)}^2\\ & =& 2{\tan }^4\theta \end{array}$

Then the area of the triangle is

$\begin{array}{rcl}∆& =& \frac{1}{2}\times 2{\tan }^4\theta \\ & =& {\tan }^4\theta \\ & =& b^4\end{array}$

Therefore, the correct statement is $1<e<\sqrt{2}$.

Hence, the correct answer is option (A) and option (D).