In the parabola y2=6 x, the equation of the chord through vertex and negative end of latus rectum, is

Question

In the parabola y2=6 x, the equation of the chord through vertex and negative end of latus rectum, is (A) y=2 x (B)  y+2 x=0 (C) y+3 x=0 (D) x+2 y=0

Tardigrade Admin · Accepted Answer

4 a=6 ⇒ a=(3/2) Negative end of latus rectum is (a,-2 a), i.e. ((3/2),-3), vertex is (0,0) Line through these two points is (y-0/x-0)=((-3-0/(3)2-0)) or 2 x+y=0

Q. In the parabola $y^{2} = 6 x$ , the equation of the chord through vertex and negative end of latus rectum, is

$y = 2 x$

$y + 2 x = 0$

$y + 3 x = 0$

$x + 2 y = 0$

$4 a = 6 \Rightarrow a = \frac{3}{2}$
Negative end of latus rectum is $(a, - 2 a)$ , i.e. $(\frac{3}{2}, - 3)$ , vertex is $(0, 0)$ Line through these two points is $\frac{y - 0}{x - 0} = (\frac{- 3 - 0}{\frac{3}{2} - 0})$ or $2 x + y = 0$

Q. In the parabola y2=6x, the equation of the chord through vertex and negative end of latus rectum, is

y=2x

y+2x=0

y+3x=0

x+2y=0