Answer:
[tex]y=-\frac{1}{4}(x-2)^{2}+9[/tex]
Step-by-step explanation:
Any point on a given parabola is equidistant from focus and directrix.
Given:
Focus of the parabola is at [tex](2,8)[/tex].
Directrix of the parabola is [tex]y=10[/tex].
Let [tex](x,y)[/tex] be any point on the parabola. Then, from the definition of a parabola,
Distance of [tex](x,y)[/tex] from focus = Distance of [tex](x,y)[/tex] from directrix.
Therefore,
[tex]\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|[/tex]
Squaring both sides, we get
[tex](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[/tex]
Hence, the equation of the parabola is [tex]y=-\frac{1}{4}(x-2)^{2}+9[/tex].
