Question

A parabola can be drawn given a focus of

Answer 1

Answer:

$y=-\frac{1}{4}(x-2)^{2}+9$

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at $(2,8)$.

Directrix of the parabola is $y=10$.

Let $(x,y)$ be any point on the parabola. Then, from the definition of a parabola,

Distance of $(x,y)$ from focus = Distance of $(x,y)$ from directrix.

Therefore,

$\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|$

Squaring both sides, we get

$(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9$

Hence, the equation of the parabola is $y=-\frac{1}{4}(x-2)^{2}+9$.