Conic Sections Question 94
Question: The equation of the tangent to the parabola $ y^{2}=4ax $ at point $ (a/t^{2},\ 2a/t) $ is
[RPET 1996]
Options:
A) $ ty=xt^{2}+a $
B) $ ty=x+at^{2} $
C) $ y=tx+at^{2} $
D) $ y=tx+(a/t^{2}) $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of the tangent to the parabola, $ y^{2}=4ax $ is $ yy_1=2a(x+x_1) $
therefore $ y.\frac{2a}{t}=2a( x+\frac{a}{t^{2}} ) $
therefore $ \frac{y}{t}=( x+\frac{a}{t^{2}} )$
therefore $ \frac{y}{t}=\frac{t^{2}x+a}{t^{2}} $
therefore $ ty=t^{2}x+a $
BETA
