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 $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$. 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 $\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

Hint: First, consider the coordinates of the point $P$. Then use the coordinates of the point $P$ and the point $\left( {1,0} \right)$ to calculate the equation of the tangent. Then using the slope of the tangent find the values of $a$ and $b$ . Substitute the values of $a$ and $b$ in the eccentricity formula to calculate the eccentricity. In the end, apply the properties and formula of the area of an isosceles triangle to get the required answer.

Formula Used: The equation of a tangent to the hyperbola passing from the point $\left( {{x_1},{y_1}} \right)$ in parametric form is: $\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$
Eccentricity of a hyperbola: $e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} $
Area of a triangle: $\dfrac{1}{2} \times base \times height$
Slope formula: The slope of a line passing through the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.

Complete step by step solution:
The given equation of a hyperbola is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, where $a > 1$ and $b < a$ .
Let $\left( {a\sec\left( u \right),b\tan\left( u \right)} \right)$ be the coordinates of the point $P$ that lies on the hyperbola.


Image: A hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ with tangent passing through the point $\left( {1,0} \right)$

The tangent is passes through the point $P\left( {a\sec\left( u \right),b\tan\left( u \right)} \right)$.
Apply the general formula of the tangent to the hyperbola in parametric form.
So, the equation of the tangent is,
$\dfrac{{x\left( {a\sec\left( u \right)} \right)}}{{{a^2}}} - \dfrac{{y\left( {b\tan\left( u \right)} \right)}}{{{b^2}}} = 1$
$ \Rightarrow \dfrac{{x\sec\left( u \right)}}{a} - \dfrac{{y\tan\left( u \right)}}{b} = 1$
Since the tangent is also passes through the point $A\left( {1,0} \right)$.
Substitute $\left( {1,0} \right)$ in above equation.
$\dfrac{{\left( 1 \right)\sec\left( u \right)}}{a} - \dfrac{{\left( 0 \right)\tan\left( u \right)}}{b} = 1$
$ \Rightarrow \dfrac{{\sec\left( u \right)}}{a} = 1$
$ \Rightarrow \sec\left( u \right) = a$
Slope of the tangent is 1.
Apply the slope formula for the tangent.
   $\dfrac{{b\tan\left( u \right) - 0}}{{a\sec\left( u \right) - 1}} = 1$
Simplify the above equation.
$\dfrac{{b\tan\left( u \right) - 0}}{{\left( {\sec\left( u \right)} \right)\sec\left( u \right) - 1}} = 1$ [since $a = \sec\left( u \right)$]
$ \Rightarrow \dfrac{{b\tan\left( u \right)}}{{\sec^{2}\left( u \right) - 1}} = 1$
$ \Rightarrow \dfrac{{b\tan\left( u \right)}}{{\tan^{2}\left( u \right)}} = 1$ [Since $\sec^{2}\left( u \right) - 1 = \tan^{2}\left( u \right)$]
$ \Rightarrow \dfrac{b}{{\tan\left( u \right)}} = 1$
$ \Rightarrow b = \tan\left( u \right)$

Substitute $a = \sec\left( u \right)$ and $b = \tan\left( u \right)$ in the formula of eccentricity.
$e = \sqrt {1 + \dfrac{{\tan^{2}\left( u \right)}}{{\sec^{2}\left( u \right)}}} $
$ \Rightarrow e = \sqrt {1 + \dfrac{{\left( {\dfrac{{\sin^{2}\left( u \right)}}{{\cos^{2}\left( u \right)}}} \right)}}{{\left( {\dfrac{1}{{\cos^{2}\left( u \right)}}} \right)}}} $ [since $\dfrac{1}{{\cos x}} = \sec x$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$]
$ \Rightarrow e = \sqrt {1 + \sin^{2}\left( u \right)} $
The range of $\sin^{2}\left( u \right)$ in first quadrant is $\left[ {0,1} \right]$.
So, the minimum value of $e$ in the first quadrant is,
$e = \sqrt {1 + \sin^{2}\left( 0 \right)} $
$ \Rightarrow e = \sqrt {1 + 0} $ [Since, $\sin\left( 0 \right) = 0$]
$ \Rightarrow e = 1$

And the maximum value of $e$ in the first quadrant is,
$e = \sqrt {1 + \sin^{2}\left( {\dfrac{\pi }{2}} \right)} $
$ \Rightarrow e = \sqrt {1 + 1} $ [Since $\sin\left( {\dfrac{\pi }{2}} \right) = 1$]
$ \Rightarrow e = \sqrt 2 $

Therefore, the eccentricity of the hyperbola lies between
$1 < e < \sqrt 2 $

Since, the vertices of the point $P$ are $\left( {\sec^{2}\left( u \right),\tan^{2}\left( u \right)} \right)$.
Apply the distance formula to calculate the distance between $A\left( {1,0} \right)$ and $P\left( {\sec^{2}\left( u \right),\tan^{2}\left( u \right)} \right)$.
$AP = \sqrt {{{\left( {\sec^{2}\left( u \right) - 1} \right)}^2} + {{\left( {\tan^{2}\left( u \right)} \right)}^2}} $
$ \Rightarrow AP = \sqrt {{{\left( {{{\tan }^2}\left( u \right)} \right)}^2} + {{\left( {\tan^{2}\left( u \right)} \right)}^2}} $
$ \Rightarrow AP = \sqrt {2\tan^{4}\left( u \right)} $
$ \Rightarrow AP = \sqrt 2 \tan^{2}\left( u \right)$

Since the $\Delta APB$ is an isosceles triangle.
So, $AP = BP$
Apply the formula of the area of a triangle.
$\Delta = \dfrac{1}{2} \times AP \times BP$
$ \Rightarrow \Delta = \dfrac{1}{2} \times {\left( {AP} \right)^2}$ [since, $AP = BP$]
$ \Rightarrow \Delta = \dfrac{1}{2} \times {\left( {\sqrt 2 \tan^{2}\left( u \right)} \right)^2}$
$ \Rightarrow \Delta = \dfrac{1}{2} \times \left( {2\tan^{4}\left( u \right)} \right)$
$ \Rightarrow \Delta = \tan^{4}\left( u \right)$
$ \Rightarrow \Delta = {b^4}$

Option ‘A’ and ‘D; are correct

Note: The equation of tangents to the hyperbola at the point is: $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$.
Two tangents can be drawn to the hyperbola from an external point: $\left( {{x_1},{y_1}} \right)$ to the hyperbola.
The equation of the normal to hyperbola at the point is: $\left( {{x_1},{y_1}} \right)$ is $\dfrac{{{a^2}x}}{{{x_1}}} - \dfrac{{{b^2}y}}{{{y_1}}} = 1$.