Analytical Geometry> let p(asec θ,btanθ) and q(asecФ,btanФ) wh...

let p(asecθ,btanθ) and q(asecФ,btanФ) where Ф+θ=π/2 be two points on the hyperbola x²/a²-y²/b²=1. if (h,k) is the point of intersection of the normals at P and Q then k is equal to
a)a²+b²/a
b)-a²+b²/a
c)a²+b²/b
d)-a²+b²/b
please answer with solution asap.

supraja venkatraman , 7 Years ago

Grade 11

1 Answers

Chakshu Shah

Last Activity: 7 Years ago

The equation of parabola is

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

the equation of the normal at a point p(asecθ,btanθ) is given as

$\frac{by}{\tan \theta }+\frac{ax}{\sec \theta }=a^{2}+b^{2}$

$b\cot \theta y+a\cos \theta x-\left ( a^{2}+b^{2} \right )=0.............\left ( i \right )$ the equation of the normal at a point q(asecФ,btanФ) is given by $\frac{by}{\tan \phi }+\frac{ax}{\sec \phi }=a^{2}+b^{2}$
$b\cot\phi y+a\cos\phi x-\left ( a^{2}+b^{2} \right )=0.............\left ( ii \right )$ solving (i) and (ii) we get $\frac{x}{-\left ( a^{2}+b^{2} \right )b\cot \theta +\left (a^{2}+b^{2} \right ) b\cot\phi }=\frac{y}{-\left (a^{2}+b^{2} \right )a\cos \phi +\left (a^{2}+b^{2} \right )a\cos\theta }$ $=\frac{1}{b\cot \theta a\cos \phi -b\cot\phi a\cos\theta }$
$\frac{y}{-\left ( a^{2}+b^{2} \right )a\cos \phi +\left ( a^{2}+b^{2} \right )a\cos \theta }$ $=\frac{1}{b\cot \theta a\cos \phi -b\cot \phi a\cos \theta }$
$=\frac{y}{a\left ( a^{2} +b^{2}\right ) \left [ \cos \theta -\cos \phi \right ]}=\frac{1}{ab\left [ \cot \theta a\cos \phi -\cot \phi a\cos \theta \right ]}$
$y= \frac{\left ( a^{2}+b^{2} \right )}{b}\left [ \frac{\cos \theta -\cos \phi }{\cot \theta a\cos \phi -\cot \phi a\cos \theta } \right ]$
$y= \frac{\left ( a^{2}+b^{2} \right )}{b}\left [ \frac{\cos \theta -\cos \left ( \frac{\pi }{2}-\theta \right ) }{\cot \theta \cos \left ( \frac{\pi }{2}-\theta \right ) -\cot \left ( \frac{\pi }{2}-\theta \right )\cos \theta } \right ]$ $\left [ \theta +\phi = \frac{\pi }{2} \right ]$
$y= \frac{\left ( a^{2}+b^{2} \right )}{b}\left [ \frac{\cos \theta -\sin \theta }{\cot \theta \sin \theta -\tan \theta \cos \theta } \right ]$
$y= \frac{\left ( a^{2}+b^{2} \right )}{b}\left [ \frac{\cos \theta -\sin \theta }{\cos \theta -\sin \theta }\right]$
$y= \frac{\left ( a^{2}+b^{2} \right )}{b}$ As (h,k) is the point of intersection so $k= \frac{\left ( a^{2}+b^{2} \right )}{b}$