What is the foci (y+3)^2=sqrt(48)(x-5) ?

The foci (y+3)^2=sqrt(48)(x-5) is (5+sqrt(3),-3)

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foci (y+3)^2=sqrt(48)(x-5)

foci

(y + 3)^{2} = 48 (x - 5)

(5 + 3, - 3)

Solution steps

Rewrite

(y + 3)^{2} = 48 (x - 5)

in the standard form:

43 (x - 5) = (y - (- 3))^{2}

(h, k) = (5, - 3), p = 3

Parabola is symmetric around the x-axis and so the focus lies a distance

p

from the center

(5, - 3)

along the x-axis

(5 + p, - 3)

= (5 + 3, - 3)

x^{2} - 7 x + 12

x^{2} - 6 y = 0

y = + 4 - x^{2}

y = \frac{9 x ^{2}}{2}

- (x + 1)^{2} + 4

What is the foci (y+3)^2=sqrt(48)(x-5) ?
The foci (y+3)^2=sqrt(48)(x-5) is (5+sqrt(3),-3)