What is the standard form of the equation of the parabola with a focus at (6,5) and a directrix of $y = - 1$ $y= -1$ ?

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Answer 1 · 2018-08-02T15:09:27

Parabola is $y = \frac{1}{12} {(x - 6)}^{2} + 2$ $y=1/12(x-6)^2+2$

Parabola is the locus of a point which moves so that its distance from a given point called focus and a given line called directrix is always equal.

Let the point be $(x, y)$ $(x,y)$ . Its distance from focus $(6, 5)$ $(6,5)$ is

$\sqrt{{(x - 6)}^{2} + {(y - 5)}^{2}}$ $sqrt((x-6)^2+(y-5)^2)$

and its distance from directrix $y = - 1$ $y=-1$ is $y + 1$ $y+1$

Hence equation of parabola is $\sqrt{{(x - 6)}^{2} + {(y - 5)}^{2}} = y + 1$ $sqrt((x-6)^2+(y-5)^2)=y+1$

and squaring ${(x - 6)}^{2} + {(y - 5)}^{2} = {(y + 1)}^{2}$ $(x-6)^2+(y-5)^2=(y+1)^2$

i.e. $x^{2} - 12 x + 36 + y^{2} - 10 y + 25 = y^{2} + 2 y + 1$ $x^2-12x+36+y^2-10y+25=y^2+2y+1$

i.e. $x^{2} - 12 x + 60 = 12 y$ $x^2-12x+60=12y$

or $12 y = {(x - 6)}^{2} + 24$ $12y=(x-6)^2+24$

or $y = \frac{1}{12} {(x - 6)}^{2} + 2$ $y=1/12(x-6)^2+2$

graph{(x^2-12x+60-12y)((x-6)^2+(y-5)^2-0.03)(y+1)=0 [-4.08, 15.92, -1.32, 8.68]}

What is the standard form of the equation of the parabola with a focus at (6,5) and a directrix of y= -1y=−1y= -1?