If (0, 4) and (0, 2) are the vertex and focus | vedclass

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

Question

(C) The vertex is $V(0, 4)$ and the focus is $S(0, 2)$.Since both points lie on the $y$-axis,the axis of the parabola is the $y$-axis.The distance bet

Vedclass · Accepted Answer

$x^2 + 8y = 32$

Vedclass · Answer

(C) The vertex is $V(0, 4)$ and the focus is $S(0, 2)$.Since both points lie on the $y$-axis,the axis of the parabola is the $y$-axis.The distance between the vertex and the focus is $a = |4 - 2| = 2$.Since the focus is below the vertex,the parabola opens downwards.The standard equation for a parabola with vertex $(h, k)$ opening downwards is $(x - h)^2 = -4a(y - k)$.Substituting $h = 0, k = 4$,and $a = 2$,we get:$(x - 0)^2 = -4(2)(y - 4)$$x^2 = -8(y - 4)$$x^2 = -8y + 32$$x^2 + 8y = 32$.

If $(0, 4)$ and $(0, 2)$ are the vertex and focus of a parabola respectively,then what is its equation?

Similar Questions

If the parabola $x^2=4ay, (a>0)$ makes an intercept of length $\sqrt{40}$ units on the line $y=1+2x$,then $4a=$

If a parabola has a horizontal axis and passes through the points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$,then the $y$-coordinate of the focus of that parabola is:

What is the equation of the pair of tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 12x$?

The length of the normal chord of the parabola $y^2 = 4x$,which makes an angle of $\frac{\pi}{4}$ with the $x$-axis,is:

The parametric equations $x - 2 = t^2$ and $y = 2t$ represent which parabola?

Vedclass Test Series

Exam Paper Generator

Online Exam Module