equation of a parabola whose vertex is (2,-3)axis is parallel to the x axis and latusrectum 8 is

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Dear Student,
$h e r e 4 a = 8 s o a = 2 h e n c e e q u a t i o n w i l l b e {(y - k)}^{2} = 4 a (x - h) {(y - (- 3))}^{2} = 8 (x - 2) s i m p l i f y t h i s t o g e t t h e a n s w e r$
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Similar questions

Q. Observe the following Lists:

List-l	List - II
A. Length of the latusrectum of the parabola $3 y^{2} = - 4 x$	1. $(5, 2)$
B.Equation of the parabola whose focus $($ 0,4 $)$ and directrix $x + 4 = 0$ is	2. $\frac{- 4}{3}$
C. The focus of the parabola $(y - 2)^{2} = 8 (x - 3)$ is	3. $\frac{4}{3}$
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	5. $x^{2} = 16 y$

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Q. Prove that the equation

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represents a parabola, whose axis is parallel to the axis of x, and find its vertex and the equation to its latus rectum.

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Equation of the parabola whose axis is $y = x$ , distance from origin to vertex is $\sqrt{2}$ and distance from origin to focus is $2 \sqrt{2},$ is (Focus and vertex lie in 1st quadrant) :

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equation of a parabola whose vertex is (2,-3)axis is parallel to the x axis and latusrectum 8 is

Dear Student, here 4a=8 so a=2 hence equation will bey-k2=4a(x-h)y--32=8x-2simplify this to get the answer Regards

Dear Student,
$h e r e 4 a = 8 s o a = 2 h e n c e e q u a t i o n w i l l b e {(y - k)}^{2} = 4 a (x - h) {(y - (- 3))}^{2} = 8 (x - 2) s i m p l i f y t h i s t o g e t t h e a n s w e r$
Regards