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) Let the vertex be $A(4, -1)$ and the focus be $S(4, -3)$.The axis of the parabola is the line passing through the focus and the vertex,which is $x

Vedclass · Accepted Answer

$x^{2} - 8x + 8y + 24 = 0$

Vedclass · Answer

(A) Let the vertex be $A(4, -1)$ and the focus be $S(4, -3)$.The axis of the parabola is the line passing through the focus and the vertex,which is $x = 4$.Since the vertex $A(4, -1)$ is at a distance $a = |-1 - (-3)| = 2$ from the focus $S(4, -3)$,and the parabola opens downwards (as the focus is below the vertex),the equation of the parabola is of the form $(x - h)^{2} = -4a(y - k)$,where $(h, k)$ is the vertex.Here,$(h, k) = (4, -1)$ and $a = 2$.Substituting these values,we get:$(x - 4)^{2} = -4(2)(y - (-1))$$(x - 4)^{2} = -8(y + 1)$$x^{2} - 8x + 16 = -8y - 8$$x^{2} - 8x + 8y + 24 = 0$.

Find the equation of the parabola with focus $(4, -3)$ and vertex $(4, -1)$.

Similar Questions

Let $A$ and $B$ be two distinct points on the parabola $y^{2}=4x$. If the axis of the parabola touches a circle of radius $r$ having $AB$ as its diameter, then the slope of the line $AB$ is

The locus of the point of intersection of the perpendicular tangents to the curve $y^2 + 4y - 6x - 2 = 0$ is:

An equilateral triangle is inscribed in the parabola $y^2 = 4ax$ such that one of its vertices is at the origin $(0, 0)$ and the other two vertices lie on the parabola. The length of its side is equal to

If the normal drawn at the point $P(9, 9)$ on the parabola $y^2 = 9x$ meets the parabola again at $Q(a, b)$,then $2a + b =$

The locus of the mid-point of the line segment joining the focus and any point on the parabola $y^{2}=4ax$ is a parabola. Find the equation of its directrix.

Vedclass Test Series

Exam Paper Generator

Online Exam Module