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
Correct Answer: A
Solution :
Solving for \[{{x}^{2}},\ {{y}^{2}}\];\[\left( \sqrt{\frac{b'-b}{ab'-ba'}}\text{,}\,\text{ }\sqrt{\frac{a'-a}{a'b-b'a}} \right)\] is the intersecting point. Differentiating\[a{{x}^{2}}+b{{y}^{2}}=1\], \[2ax+2by\frac{dy}{dx}=0\] \[\Rightarrow {{\left( \frac{dy}{dx} \right)}_{1}}=-\frac{ax}{ay}\] and \[{{\left( \frac{dy}{dx} \right)}_{2}}=-\frac{a'x}{b'y}\] and \[{{\left( \frac{dy}{dx} \right)}_{1}}{{\left( \frac{dy}{dx} \right)}_{2}}=-1\] \[\Rightarrow \frac{aa'}{bb'}\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)=-1\]or \[\frac{aa'}{bb'}\left( \frac{b'-b}{a'-a} \right)=1\]. Hence\[\frac{1}{b}-\frac{1}{b'}=\left( \frac{1}{a}-\frac{1}{a'} \right)\].
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