Circle And System Of Circles Question 139
Question: The condition of the curves $ ax^{2}+by^{2}=1 $ and $ a’x^{2}+b’y^{2}=1 $ to intersect each other orthogonally, is
Options:
A) $ \frac{1}{a}-\frac{1}{a’}=\frac{1}{b}-\frac{1}{b’} $
B) $ \frac{1}{a}+\frac{1}{a’}=\frac{1}{b}+\frac{1}{b’} $
C) $ \frac{1}{a}+\frac{1}{b}=\frac{1}{a’}+\frac{1}{b’} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Solving for $ x^{2},\ y^{2} $ ; $ ( \sqrt{\frac{b’-b}{ab’-ba’}}\text{,},\sqrt{\frac{a’-a}{a’b-b’a}} ) $ is the intersecting point.
Differentiating $ ax^{2}+by^{2}=1 $ , $ 2ax+2by\frac{dy}{dx}=0 $
$ \Rightarrow {{( \frac{dy}{dx} )}_1}=-\frac{ax}{ay} $ and $ {{( \frac{dy}{dx} )}_2}=-\frac{a’x}{b’y} $ and $ {{( \frac{dy}{dx} )}_1}{{( \frac{dy}{dx} )}_2}=-1 $
$ \Rightarrow \frac{aa’}{bb’}( \frac{x^{2}}{y^{2}} )=-1 $ or $ \frac{aa’}{bb’}( \frac{b’-b}{a’-a} )=1 $ .
Hence $ \frac{1}{b}-\frac{1}{b’}=( \frac{1}{a}-\frac{1}{a’} ) $ .
BETA
