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Question: Let \[P\left( {asec\theta ,btan\theta } \right)\] and \[Q\left( {asec\phi ,btan\phi } \right),\] whe...

Let P(asecθ,btanθ)P\left( {asec\theta ,btan\theta } \right) and Q(asecϕ,btanϕ),Q\left( {asec\phi ,btan\phi } \right), where θ+Φ=π2\theta + \Phi = \dfrac{\pi }{2} , be two points on the hyperbola x2a2−y2b2=1\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 . If (h,k)(h,k) is the point of intersection of the normal at P & Q, then k is equal to
(A) a2+b2a\dfrac{{{a^2} + {b^2}}}{a}
(B) −(a2+b2a) - \left( {\dfrac{{{a^2} + {b^2}}}{a}} \right)
(C) a2+b2b\dfrac{{{a^2} + {b^2}}}{b}
(D) −(a2+b2b) - \left( {\dfrac{{{a^2} + {b^2}}}{b}} \right)

Explanation

Solution

The equation of normal at (asec⁡θ,btan⁡θ)(a\sec \theta ,b\tan \theta ) to the hyperbola x2a2−y2b2=1\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 is axcos⁡θ+bycotθ=a2+b2ax\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 ahcos⁡θ+bkcot⁡θ=a2+b2ah\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 (asec⁡θ,btan⁡θ)(a\sec \theta ,b\tan \theta ) to the hyperbola x2a2−y2b2=1\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 isaxcos⁡θ+bycotθ=a2+b2ax\cos \theta + bycot\theta = {a^2} + {b^2} .
Therefore, normal at θ,Φ\theta ,\Phi are
axcos⁡θ+bycot⁡Φ=a2+b2ax\cos \theta + by\cot \Phi = {a^2} + {b^2}
axcos⁡Φ+bycot⁡Φ=a2+b2ax\cos \Phi + by\cot \Phi = {a^2} + {b^2}
Given, θ+Φ=π2\theta + \Phi = \dfrac{\pi }{2}
Therefore, Φ=π2−θ\Phi = \dfrac{\pi }{2} - \theta and these passes through (h,k)
Therefore, ahcos⁡θ+bkcot⁡θ=a2+b2.........(1)ah\cos \theta + bk\cot \theta = {a^2} + {b^2}.........(1)
Multiplying equation (1) by sin⁡θ\sin \theta , we get
ahCos⁡θSin⁡θ+bkCos⁡θ=(a2+b2)Sin⁡θah\operatorname{Cos} \theta \operatorname{Sin} \theta + bk\operatorname{Cos} \theta = ({a^2} + {b^2})\operatorname{Sin} \theta
And,
ahsin⁡θ+bktan⁡θ=a2+b2...........(2)ah\sin \theta + bk\tan \theta = {a^2} + {b^2}...........(2)
Multiplying equation (2) by Cos⁡θ\operatorname{Cos} \theta , we get
ahSin⁡θCos⁡θ+bktan⁡θCos⁡θ=(a2+b2)Cos⁡θah\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),
⇒(bk+a2+b2)(Sin⁡θ−Cos⁡θ)=0 ⇒k=−(a2+b2b) \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 x2a2−y2b2=1\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 (x1,y1)({x_1},{y_1}) is y−y1=−y1a2x1b2(x−x1),x1≠0y - {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 (asec⁡θ,btan⁡θ)(a\sec \theta ,b\tan \theta ) is axcos⁡θ+bycotθ=a2+b2ax\cos \theta + bycot\theta = {a^2} + {b^2}