Conic Sections Question 28
Question: If the two tangents drawn on hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ in such a way that the product of their gradients is $ c^{2} $ , then they intersects on the curve
Options:
A) $ y^{2}+b^{2}=c^{2}(x^{2}-a^{2}) $
B) $ y^{2}+b^{2}=c^{2}(x^{2}+a^{2}) $
C) $ ax^{2}+by^{2}=c^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ (h,k) $ be the point of intersection. By $ SS_1=T^{2} $ ,
$ ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-1 )( \frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}-1 )={{[ \frac{hx}{a^{2}}-\frac{ky}{b^{2}}-1 ]}^{2}} $
therefore $ x^{2}[ \frac{h^{2}}{a^{4}}-\frac{k^{2}}{a^{2}b^{2}}-\frac{1}{a^{2}}-\frac{h^{2}}{a^{4}} ]-y^{2}[ \frac{h^{2}}{a^{2}b^{2}}-\frac{k^{2}}{b^{4}}-\frac{1}{b^{2}}+\frac{k^{2}}{b^{4}} ]+…=0 $
We know that, $ m_1m_2=\frac{Coefficentofx^{2}}{Coefficentofy^{2}} $
$ \Rightarrow $ $ m_1m_2=\frac{\frac{k^{2}}{a^{2}b^{2}}+\frac{1}{a^{2}}}{\frac{h^{2}}{a^{2}b^{2}}-\frac{1}{b^{2}}}=c^{2} $
$ \Rightarrow $ $ ( \frac{k^{2}+b^{2}}{h^{2}-a^{2}} )=c^{2} $ or $ (y^{2}+b^{2})=c^{2}(x^{2}-a^{2}) $ .
BETA
