Question

Let \[P\left( {asec\theta ,btan\theta } \right)\] and \[Q\left( {asec\phi ,btan\phi } \right),\] where $\theta + \Phi = \dfrac{\pi }{2}$ , be two points on the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ . If $(h,k)$ is the point of intersection of the normal at P & Q, then k is equal to
(A) $\dfrac{{{a^2} + {b^2}}}{a}$
(B) $ - \left( {\dfrac{{{a^2} + {b^2}}}{a}} \right)$
(C) $\dfrac{{{a^2} + {b^2}}}{b}$
(D) $ - \left( {\dfrac{{{a^2} + {b^2}}}{b}} \right)$

Answer 1

Hint:The equation of normal at $(a\sec \theta ,b\tan \theta )$ to the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $ax\cos \theta + bycot\theta = {a^2} + {b^2}$ . If (h,k) the point of intersection of the normal. Then in the place x,y we will put (h,k). The equation of normal becomes $ah\cos \theta + bk\cot \theta = {a^2} + {b^2}$.

Complete step-by-step answer:

The normal of the Hyperbola is HK.
The equation of normal at $(a\sec \theta ,b\tan \theta )$ to the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is$ax\cos \theta + bycot\theta = {a^2} + {b^2}$ .
Therefore, normal at $\theta ,\Phi $ are
$ax\cos \theta + by\cot \Phi = {a^2} + {b^2}$
$ax\cos \Phi + by\cot \Phi = {a^2} + {b^2}$
Given, $\theta + \Phi = \dfrac{\pi }{2}$
Therefore, $\Phi = \dfrac{\pi }{2} - \theta $ and these passes through (h,k)
Therefore, $ah\cos \theta + bk\cot \theta = {a^2} + {b^2}.........(1)$
Multiplying equation (1) by $\sin \theta $ , we get
$ah\operatorname{Cos} \theta \operatorname{Sin} \theta + bk\operatorname{Cos} \theta = ({a^2} + {b^2})\operatorname{Sin} \theta $
And,
$ah\sin \theta + bk\tan \theta = {a^2} + {b^2}...........(2)$
Multiplying equation (2) by $\operatorname{Cos} \theta $ , we get
$ah\operatorname{Sin} \theta \operatorname{Cos} \theta + bk\tan \theta \operatorname{Cos} \theta = ({a^2} + {b^2})\operatorname{Cos} \theta $
Subtract, Both the equations (1) and (2),
$
   \Rightarrow (bk + {a^2} + {b^2})(\operatorname{Sin} \theta - \operatorname{Cos} \theta ) = 0 \\
   \Rightarrow k = - \left( {\dfrac{{{a^2} + {b^2}}}{b}} \right)$

So, the correct answer is “Option D”.

Note:

The Normal is HK.
For the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, The equation of normal in different forms are-
In the cartesian form equation of normal at point \[({x_1},{y_1})\] is $y - {y_1} = \dfrac{{ - {y_1}{a^2}}}{{{x_1}{b^2}}}(x - {x_1}),{x_1} \ne 0$
In the parametric form, equation of normal at $(a\sec \theta ,b\tan \theta )$ is $ax\cos \theta + bycot\theta = {a^2} + {b^2}$