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

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.

Grade:11

1 Answers

Chakshu Shah
33 Points
5 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}

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