Conic Sections Question 147
Question: The value of m for which $ y=mx+6 $ is a tangent to the hyperbola $ \frac{x^{2}}{100}-\frac{y^{2}}{49}=1 $ , is
[Karnataka CET 1993]
Options:
A) $ \sqrt{\frac{17}{20}} $
B) $ \sqrt{\frac{20}{17}} $
C) $ \sqrt{\frac{3}{20}} $
D) $ \sqrt{\frac{20}{3}} $
Show Answer
Answer:
Correct Answer: A
Solution:
If $ y=mx+c $ touches $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1, $
then $ c^{2}=a^{2}m^{2}-b^{2} $ . Here $ c=6,a^{2}=100,b^{2}=49 $
$ \therefore 36=100m^{2}-49\Rightarrow 100m^{2}=85\Rightarrow m=\sqrt{\frac{17}{20}} $ .
BETA
