A parabola can be drawn given a focus of (11, 8) and a directrix of x = 1. Write theequation of the parabola in any form. I’m

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Answer:

[tex](y-8)^2=20(x-6)[/tex]

Explanation:

Note that any point on the parabola is equidistant from the focus and directrix.

Distance from the point (x, y) to the directrix x = 1 is (x-1)

Also, distance from the point (x, y) to the focus (11,8) is expressed as;

[tex]\begin{gathered} L^2=(x-11)^2+(y-8)^2 \\ Since\text{ L = x-1} \\ \text{Substitute;} \\ (x-1)^2=(x-11)^2+(y-8)^2 \\ x^2-2x+1=x^2-22x+121+(y-8)^2 \\ x^2-x^2-2x+22x+1-121\text{ = }(y-8)^2 \\ 20x-120\text{ = }(y-8)^2 \\ 20(x-6)=(y-8)^2 \\ \text{Swap} \\ (y-8)^2=20(x-6) \end{gathered}[/tex]

Hence the required equation of the parabola in any form is expressed as;

[tex](y-8)^2=20(x-6)[/tex]

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