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skad [1K]
2 years ago
8

Write an equation of a parabola that opens to the left, has a vertex at the origin, and a focus at (–4, 0).

Mathematics
1 answer:
8_murik_8 [283]2 years ago
8 0

Answer:

y^{2}=-16x

Step-by-step explanation:

we know that

The standard equation of a horizontal parabola is equal to  

(y-k)^{2}=4p(x-h)

where

(h,k) is the vertex

(h+p,k) is the focus

In this problem we have

(h,k)=(0,0) ----> vertex at origin

(h+p,k)=(-4,0)

so

h+p=-4

p=-4

substitute the values

(y-0)^{2}=4(-4)(x-0)

y^{2}=-16x

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<u>Solution-</u>

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

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=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t) = 13.5191




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