Conic Sections Question 137
Question: In the parabola $ y^{2}=6x $ , the equation of the chord through vertex and negative end of latus rectum, is
Options:
A) $ y=2x $
B) $ y+2x=0 $
C) $ x=2y $
D) $ x+2y=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Vertex $ \equiv (0,0). $ End points of latus rectum are $ (a,\pm 2a) $ . Here $ a=\frac{6}{4} $
Therefore, $ -ve $ end of latus rectum is $ ( \frac{3}{2},-3 ) $
Line through the point is $ y=\frac{-3}{3/2}x $ or $ y+2x=0 $ .
BETA
