Conic-Sections Question 540
Question: If $ {{( \frac{x}{a} )}^{2}}+( {{\frac{y}{b}}^{2}} )=1(a>b) $ and $ x^{2}-y^{2}=c^{2} $ cut at right angles, then
Options:
A) $ a^{2}+b^{2}=2c^{2} $
B) $ b^{2}-a^{2}=2c^{2} $
C) $ a^{2}-b^{2}=2c^{2} $
D) $ a^{2}-b^{2}=c^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ (x_1,y_1) $ be their point of intersection then $ x^2_1-y^2_1=c^{2} $ ? (1) $ \frac{x^2_1}{a^{2}}+\frac{y^2_1}{b^{2}}=1 $ ? (2)
$ \Rightarrow x^2_1( \frac{1}{a^{2}}-\frac{1}{c^{2}} )+y^2_1( \frac{1}{b^{2}}+\frac{1}{c^{2}} )=0 $ ? (3) Now tangents to the curves are $ xx_1-yy_1=c^{2} $ And $ \frac{xx_1}{a^{2}}+\frac{yy_1}{b^{2}}=1 $ The tangents are perpendicular, so $ \frac{x_1}{y_1}\times -\frac{b^{2}}{a^{2}}\frac{x_1}{y_1}=-1\Rightarrow b^{2}x^2_1-a^{2}y^2_1=0 $ ? (4) Eliminating $ x^2_1 $ and $ y^2_1 $ from (3) and (4) we get, $ \frac{c^{2}-a^{2}}{a^{2}b^{2}c^{2}}=\frac{-(b^{2}+c^{2})}{a^{2}b^{2}c^{2}}\Rightarrow a^{2}-b^{2}=2c^{2} $
BETA
