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Derive the equation of the parabola with a focus at (_5, 5) and a directrix of y = -1.


f(x) = _one twelfth (x _ 5)2 + 2

f(x) = one twelfth (x _ 5)2 + 2

f(x) = _one twelfth (x + 5)2 + 2

f(x) = one twelfth (x + 5)2 + 2

Verified Answer :

Answer: Option D is correct.

Step-by-step explanation:

Since we have given that

focus = (-5,5)

and a directrix y= -1

Since, equation of parabola in this case will be

\((x-h)^2=4.a(y-k)\)

Now, here

\(y=k-a=-1\\\\\text { focus =(h,k+a)}\\\\\text{So,} k+a=5\\\\\text{ by solving these two equation , we get }\\\\a=3\text{ and } k=2\)

So equation will be

\((x+5)^2=4\times 3(y-2)\\\\(x+5)^2=12(y-3)\\\\y=\frac{1}{12}(x+5)^2+2\)

So, option D is correct .

zame

Answer:

D)  f(x) = one twelfth(x + 5)2 + 2

Step-by-step explanation:

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