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  • Lwt <img alt="P(a \sec \theta, b \tan \theta) \text { and } Q(a \sec \phi, b \tan \phi)" src="https://entrancecorner.oncodecogs.com/gif.latex?P%28a%20%5Csec%20%5Ctheta%2C%20b%20%5Ctan%20%

Lwt P(a \sec \theta, b \tan \theta) \text { and } Q(a \sec \phi, b \tan \phi) where  \theta+\phi=\frac{\pi}{2}  be two points on the hyperbola  \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 . If (h,k) is the point of intersection of the normal P and Q, then k is equal to

Option: 1

\begin{aligned} & \frac{a^2+b^2}{a} \\ \end{aligned}


Option: 2

-\left(\frac{a^2+b^2}{a}\right) \\


Option: 3

\left(\frac{a^2+b^2}{b}\right) \\


Option: 4

-\left(\frac{a^2+b^2}{b}\right)


Answers (1)

best_answer

Equation of the parabola is   \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, at the point  (a \sec \alpha, b \tan \alpha) is given by  a x \cos \alpha+b y \cot \alpha=a^2+b^2

Normal at   \theta, \phi \text { are } \quad\left\{\begin{array}{l} a x \cos \theta+b y \cot \theta=a^2+b^2 \\ a x \cos \phi+b y \cot \phi=a^2+b^2 \end{array}\right.

Where  \theta+\phi=\frac{\pi}{2}, and these passes through (h,k)

\begin{gathered} \therefore a h \cos \theta+b k \cot \theta=a^2+b^2 \\ a h \sin \theta+b k \tan \theta=a^2+b^2 \end{gathered}

Eliminating h, 

\begin{aligned} & b k(\cot \theta \sin \theta-\tan \theta \cos \theta)=\left(a^2+b^2\right)(\sin \theta-\cos \theta) \\ & k=-\left(\frac{a^2+b^2}{b}\right) \end{aligned}

 

Posted by

jitender.kumar

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