Conic Sections Question 235
Question: The angle of intersection of ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ and circle $ x^{2}+y^{2}=ab $ , is
Options:
A) $ {{\tan }^{-1}}( \frac{a-b}{ab} ) $
B) $ {{\tan }^{-1}}( \frac{a+b}{ab} ) $
C) $ {{\tan }^{-1}}( \frac{a+b}{\sqrt{ab}} ) $
D) $ {{\tan }^{-1}}( \frac{a-b}{\sqrt{ab}} ) $
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Answer:
Correct Answer: D
Solution:
$ \frac{ab-y^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ or $ y^{2}( \frac{a^{2}-b^{2}}{a^{2}b^{2}} )=\frac{a-b}{a} $
or $ y^{2}=( \frac{ab^{2}}{a+b} ) $ and $ x^{2}=( \frac{a^{2}b}{a+b} )\Rightarrow (x,y)=( a\sqrt{\frac{b}{a+b}},b\sqrt{\frac{a}{a+b}} ) $
Slope of tangent at ellipse $ =\frac{-b^{2}x}{a^{2}y}=\frac{-b^{2}}{a^{2}}\sqrt{\frac{a}{b}} $
Slope of tangent at circle $ =-\frac{x}{y}=-\sqrt{\frac{a}{b}} $
$ \therefore \theta ={{\tan }^{-1}}[ \frac{-\sqrt{\frac{a}{b}}+\frac{b^{2}}{a^{2}}\sqrt{\frac{a}{b}}}{1+\frac{b^{2}}{a^{2}}.\frac{a}{b}} ] $ i.e., $ \theta ={{\tan }^{-1}}[ \frac{a-b}{\sqrt{ab}} ] $ . Note : Students should remember this question as a formula.
BETA
