Conic-Sections Question 596
Question: The locus of the point of intersection of two tangents of the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ which are inclined at angles $ {\theta_1} $ , and $ {\theta_2} $ with the major axis such that $ {{\tan }^{2}}{\theta_1}+{{\tan }^{2}}{\theta_2} $ is constant, is
Options:
A) $ 4x^{2}y^{2}+2(x^{2}-a^{2})(y^{2}-b^{2})=k{{(x^{2}-a^{2})}^{2}} $
B) $ 4x^{2}y^{2}-2(x^{2}-a^{2})(y^{2}-b^{2})=k{{(x^{2}-a^{2})}^{2}} $
C) $ 4x^{2}y^{2}-2(x^{2}-a^{2})(y^{2}-b^{2})=k{{(x^{2}+a^{2})}^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] If the tangent $ y=mx\pm \sqrt{a^{2}m^{2}+b^{2}} $ passes through $ (x_1,y_1) $ then $ {{(y_1-mx_1)}^{2}}=a^{2}m^{2}+b^{2} $ or $ (x^2_1-a^{2})m^{2}-2x_1y_1m+y^2_1-b^{2}=0. $ If The roots be $ m_1 $ and $ m_2 $ then $ m_1+m_2=\frac{2x_1y_1}{x^2_1-a^{2}} $ and $ m_1m_2=\frac{y^2_1-b^{2}}{x^2_1-a^{2}} $ Given: $ {{\tan }^{2}}{\theta_1}+{{\tan }^{2}}{\theta_2}= $ constant $ =k $ (say)
$ \Rightarrow m^2_1+m^2_2=k $ or $ {{(m_1+m_2)}^{2}}-2m_1m_2=k $
$ \Rightarrow \frac{4x^{2}y^{2}}{{{(x^2_1-a^{2})}^{2}}}-\frac{2(y^2_1-b^{2})}{(x^2_1-a^{2})}=k $ or $ 4x^2_1y^2_1-2(x^2_1-a^{2})(y^2_1-b^{2})=k{{(x^2_1-a^{2})}^{2}} $ Hence locus of P is $ 4x^{2}y^{2}-2(x^{2}-a^{2})(y^{2}-b^{2})=k{{(x^{2}-a^{2})}^{2}} $
BETA
