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 $(0, -3)$ and the directrix is $y = 3$.Since the focus lies on the $y$-axis and the directrix is a horizontal line,th

Vedclass · Accepted Answer

$x^2 = -12y$

Vedclass · Answer

(A) The focus of the parabola is $(0, -3)$ and the directrix is $y = 3$.Since the focus lies on the $y$-axis and the directrix is a horizontal line,the parabola is symmetric about the $y$-axis.The standard form of such a parabola is $x^2 = 4ay$.The focus is $(0, a)$ and the directrix is $y = -a$.Comparing $(0, -3)$ with $(0, a)$,we get $a = -3$.Substituting $a = -3$ into the equation $x^2 = 4ay$,we get $x^2 = 4(-3)y$,which simplifies to $x^2 = -12y$.

Find the equation of the parabola with focus $(0, -3)$ and directrix $y = 3$.

Similar Questions

If $2x + 3y + 12 = 0$ and $x - y + 4\lambda = 0$ are conjugate with respect to the parabola $y^2 = 8x$,then $\lambda$ is equal to

Let $ABCD$ be a trapezium whose vertices lie on the parabola $y^2=4x$. Let the sides $AD$ and $BC$ of the trapezium be parallel to the $y$-axis. If the diagonal $AC$ is of length $\frac{25}{4}$ and it passes through the point $(1,0)$,then the area of $ABCD$ is:

The equation of a straight line drawn through the focus of the parabola $y^2 = -4x$ at an angle of $120^\circ$ to the $x$-axis is:

The diameter of the parabola $y^2 = 4ax$ which bisects the chords parallel to $y = mx + \alpha$ is:

If $(2, -8)$ is one endpoint of a focal chord of the parabola $y^{2} = 32x$,what is the other endpoint?

Vedclass Test Series

Exam Paper Generator

Online Exam Module