Question

Hello can someone help me please and thanksss

Answer 1

the equation of the parabola is given as $y = 1/4 (x-6)^2 +1$. D

How to find the parabola

Given the focus(6,2) and the directrix, y = 0

Use the formula

$\sqrt{(x-6)^{2} + (y-2)^{2} } }$ = (y - 0)

Find the square of both sides, square root is removed from the side with it and square added on the other side

$(x-6)^2 + (y-2)^2 = (y- 0)^{2}$

Expand the expression and bring all terms to one side

$x^{2} - 6x - 6x + 36 +y^{2} -2y- 2y +4$ = $y^{2} - 0$

Collect like terms

$x^{2} - 12x -4y + 40 = 0$

Make 'y' the subject of the formula

$4y = x^{2} - 12x + 40$

Divide through by 4 and substrate 4 from 40 to give a perfect quadratic equation

$y = 1/4( x^{2} - 12x + 36) + 4$

1/4 is dividing the quadratic equation $x^2 - 12x +36$

Then divide factor by 4 to give 1, this is so because of the nature of the options given

$y = 1/4( x^{2} -6x -6x +36) + 1$

Simplify the expanded quadratic equation

$y = x (x -6) - 6 (x- 6)$, we have $(x - 6) ^2$

Then insert in place into previous equation

$y = 1/4 (x-6)^2 + 1$

Thus, the equation of the parabola is given as $y = 1/4 (x-6)^2 +1$

Learn more about a parabola here:

https://brainly.com/question/4061870

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