Conic Sections Question 71
Question: Let $ P(a\sec \theta ,\ b\tan \theta ) $ and $ Q(a\sec \varphi ,\ b\tan \varphi ) $ , where $ \theta +\varphi =\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 normals at P and Q, then k is equal to
[IIT 1999; MP PET 2002]
Options:
A) $ \frac{a^{2}+b^{2}}{a} $
B) $ -( \frac{a^{2}+b^{2}}{a} ) $
C) $ \frac{a^{2}+b^{2}}{b} $
D) $ -( \frac{a^{2}+b^{2}}{b} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
Given P $ (a\sec \theta ,b\tan \theta ) $ and $ Q(a\sec \varphi ,b\tan \varphi ) $
The equation of tangent at point P is $ \frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1 $
m of tangent $ =\frac{b}{\tan \theta }\times \frac{\sec \theta }{a}=\frac{b}{a}.\frac{1}{\sin \theta } $
Hence the equation of perpendicular at P is $ y-b\tan \theta =-\frac{a\sin \theta }{b}(x-a\sec \theta ) $
or $ by-b^{2}\tan \theta =-a\sin \theta x+a^{2}\tan \theta $
or $ a\sin \theta x+by=(a^{2}+b^{2})\tan \theta $ ……(i) Similarly the equation of perpendicular at Q is $ a\sin \varphi x+by=(a^{2}+b^{2})\tan \varphi $ ……(ii) On multiplying (i) by $ \sin \varphi $ and (ii) by $ \sin \theta $
$ a\sin \theta \sin \varphi x+b\sin \varphi y=(a^{2}+b^{2})\tan \theta \sin \varphi $
$ a\sin \varphi \sin \theta x+b\sin \theta y=(a^{2}+b^{2})\tan \varphi \sin \theta $
On subtraction by, $ (\sin \varphi -\sin \theta )=(a^{2}+b^{2})(\tan \theta \sin \varphi -\tan \varphi \sin \theta ) $
$ \therefore y=k=\frac{a^{2}+b^{2}}{b}.\frac{\tan \theta \sin \varphi -\tan \varphi \sin \theta }{\sin \varphi -\sin \theta } $
$ \because \theta +\varphi =\frac{\pi }{2} $
therefore $ \varphi =\frac{\pi }{2}-\theta $
therefore $ \sin \varphi =\cos \theta $ and $ \tan \varphi =\cot \theta $
$ \therefore y=k=\frac{a^{2}+b^{2}}{b}.\frac{\tan \theta \cos \theta -\cot \theta \sin \theta }{\cos \theta -\sin \theta } $
$ =\frac{a^{2}+b^{2}}{b}( \frac{\sin \theta -\cos \theta }{\cos \theta -\sin \theta } )=-\frac{(a^{2}+b^{2})}{b} $ .
BETA
