In the parabola y^2 = 6x,the equation of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the parabola is $y^2 = 6x$. Comparing this with $y^2 = 4ax$,we get $4a = 6$,so $a = \frac{3}{2}$.The vertex of the parabola is $(0

Vedclass · Accepted Answer

$y + 2x = 0$

Vedclass · Answer

(B) The equation of the parabola is $y^2 = 6x$. Comparing this with $y^2 = 4ax$,we get $4a = 6$,so $a = \frac{3}{2}$.The vertex of the parabola is $(0, 0)$.The coordinates of the endpoints of the latus rectum are $(a, 2a)$ and $(a, -2a)$.The negative end of the latus rectum is $(a, -2a) = (\frac{3}{2}, -2 	imes \frac{3}{2}) = (\frac{3}{2}, -3)$.The equation of the line passing through the vertex $(0, 0)$ and the point $(\frac{3}{2}, -3)$ is given by $y - 0 = \frac{-3 - 0}{3/2 - 0}(x - 0)$.$y = \frac{-3}{3/2}x = -2x$.Therefore,$y + 2x = 0$.

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

Similar Questions

Find the equation of the common tangent to the parabolas $y = x^2$ and $y = -(x - 2)^2$.

The length of the latus rectum of a parabola,whose vertex and focus are on the positive $x$-axis at a distance $R$ and $S$ $(S > R)$ respectively from the origin,is:

If all the vertices of an equilateral triangle lie on the parabola $y^2=16x$ and one of them coincides with the vertex of that parabola,then the length of the side of that triangle is

If the normal to the parabola ${y^2} = 12x$ at the point $(3, 6)$ meets the parabola again at the point $(27, -18)$,then the equation of the circle having this normal chord as its diameter is:

The points of intersection of the parabolas $y^2=5x$ and $x^2=5y$ lie on the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module