The equation of the parabola whose focus is at (3, 0) and the directrix is at x = –3 is y=6x²+6x.
y = a(x-h)² + k
where,
(h, k) are the coordinates of the vertex of the parabola in form (x, y);
a defines how narrower is the parabola, and the "-" or "+" that the parabola will open up or down.
As it is given that the focus of the parabola is at F(3,0) and its directrix is the line x=−3, therefore, we can write that
x=-3
x+3=0
Now, Let A(x,y) be any point in the plane of the directrix and focus, and MA be the perpendicular distance from A to the directrix, therefore, A lies on parabola if FA=MA,
[tex]FA=MA\\\\\sqrt{(x-3)^2 +(y-0)^2}=\dfrac{|x+3|}{1}\\\\{(x-3)^2 +(y-0)^2}={(x+3)^2}\\\\-6x^2+y=6x\\\\y=6x^2+6x[/tex]
Hence, the equation of the parabola whose focus is at (3, 0) and the directrix is at x = –3 is y=6x²+6x.
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