Conic Sections Question 215
Question: Tangents at any point on the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ cut the axes at A and B respectively. If the rectangle OAPB (where O is the origin)is completed, then locus of point P is given by :
Options:
A) $ \frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}=1 $
B) $ \frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=1 $
C) $ \frac{a^{2}}{y^{2}}-\frac{b^{2}}{x^{2}}=1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Eq. of the tangent at the point $ ‘\theta ’ $ is $ \frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1 $
$ \Rightarrow $ A is $ (acos\theta ,0) $ and B is $ (0,-bcot\theta ) $ Let P be $ (h,k)\Rightarrow h=acos\theta ,k=-bcot\theta $
$ \Rightarrow \frac{k}{h}=-\frac{b}{a\sin \theta }\Rightarrow \sin \theta =\frac{bh}{ak} $ and $ \cos \theta =\frac{h}{a}. $ Square and add,
$ \Rightarrow \frac{b^{2}h^{2}}{a^{2}k^{2}}+\frac{h^{2}}{a^{2}}=1\Rightarrow \frac{b^{2}}{k^{2}}+1=\frac{a^{2}}{h^{2}} $
$ \Rightarrow \frac{a^{2}}{h^{2}}-\frac{b^{2}}{k^{2}}=1 $
Hence, locus of P is $ \frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}=1 $
BETA
