Conic Sections Question 194
Question: If the circle $ x^{2}+y^{2}=a^{2} $ intersects the hyperbola $ xy=c^{2} $ in four points $ P(x_1,y_1),Q(x_2,y_2),R(x_3,y_3),S(x_4,y_4) $ Then
Options:
A) $ x_1+x_2+x_3+x_4=0 $
B) $ y_1+y_2+y_3+y_4=2 $
C) $ x_1x_2x_3x_4=2c^{4} $
D) $ y_1y_2y_3y_4=2c^{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ (x _{i},y _{i}),i=1,2,3,4 $ lies on $ xy=c^{2}\Rightarrow y _{i}=\frac{c^{2}}{x _{i}} $ Now the point $ (x _{i},y _{i}) $ lies on $ x^{2}+y^{2}=a^{2}\Rightarrow x^2 _{i}+\frac{c^{4}}{x _{i}}=a^{2} $
$ \Rightarrow x^4 _{i}-a^{2}x^2 _{i}+c^{4}=0 $ Its roots are $ x_1,x_2,x_3,x_4\therefore $ $ x_1+x_2+x_3+x_4=0 $ $ x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=a^{2} $ $ x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=0 $ $ x_1x_2x_3x_4=c^{4} $ Clearly [c] is not correct Now $ y_1y_2y_3y_4=\frac{c^{2}}{x_1}.\frac{c^{2}}{x_2}.\frac{c^{2}}{x_3}.\frac{c^{2}}{x_4}=c^{4} $ and $ y_1+y_2+y_3+y_4=\frac{c^{2}(\Sigma x_1x_2x_3)}{x_1x_2x_3x_4}=0 $
BETA
