Conic Sections Question 381
Question: The equation of the normal at the point (6, 4) on the hyperbola $ \frac{x^{2}}{9}-\frac{y^{2}}{16}=3 $ , is
Options:
A) $ 3x+8y=50 $
B) $ 3x-8y=50 $
C) $ 8x+3y=50 $
D) $ 8x-3y=50 $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of normal at any point $ (x_1,y_1) $ on hyperbola is, $ \frac{a^{2}(x-x_1)}{x_1}=\frac{b^{2}(y-y_1)}{-y_1} $
Here, $ a^{2}=267,b^{2}=48 $ and $ (x_1,y_1)=(6,4) $
$ \therefore \frac{27(x-6)}{6}=-\frac{48(y-4)}{4} $
therefore $ 3(x-6)=-8(y-4) $
therefore $ 3x+8y=50 $ .
BETA
