Find the equation of the parabola with focus | vedclass

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

Question

(A) The focus of the parabola is $(2, 0)$,which lies on the $x$-axis. Therefore,the $x$-axis is the axis of the parabola.Since the directrix is $x = -

Vedclass · Accepted Answer

$y^{2} = 8x$

Vedclass · Answer

(A) The focus of the parabola is $(2, 0)$,which lies on the $x$-axis. Therefore,the $x$-axis is the axis of the parabola.Since the directrix is $x = -2$ (which is of the form $x = -a$),the parabola opens to the right and is of the form $y^{2} = 4ax$.Here,the distance from the vertex $(0, 0)$ to the focus $(2, 0)$ is $a = 2$.Substituting $a = 2$ into the standard equation $y^{2} = 4ax$,we get $y^{2} = 4(2)x = 8x$.

Find the equation of the parabola with focus $(2, 0)$ and directrix $x = -2$.

Similar Questions

The equation of the parabola with focus $(0,0)$ and directrix $x+y=4$ is

If the line $y=2x+k$ is a normal to the parabola $y^2=4x$,then $k=$

If the normal chord drawn at $(2 a, 2 a \sqrt{2})$ on the parabola $y^2=4 a x$ subtends an angle $\theta$ at its vertex,then $\theta=$ (in $^{\circ}$)

The vertex of the parabola is at $(1, 2)$ and its axis is parallel to the $y$-axis. If the parabola passes through $(0, 6)$,then its latus rectum is:

The equation of a tangent to the parabola,$x^2 = 8y,$ which makes an angle $\theta$ with the positive direction of the $x-$axis,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module