How do you find the standard form of the equation of the parabola with focus (7, 5) & directrix x=1?

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Answer 1 · 2016-07-06T06:08:52

$(y-5)^2=12(x-4)$ is the reqd. eqn.

Let $S(7,5)$ be the Focus & the eqn. of Dir. $d: x-1=0.$

If $P(x,y)$ be any point on the reqd. Parabola , then by the Focus-Directrix Property of Parabola, we have,

Dist. $SP$ = Perp. Dist. btwn. $(P,d).$

$:. sqrt[(x-7)^2+(y-5)^2]=|x-1|.................(1)$

$:.(x-7)^2+(y-5)^2=|x-1|^2$

$:. (y-5)^2=x^2-2x+1-x^2+14x-49=12x-48,$ i.e.,

$(y-5)^2=12(x-4)$ is the reqd. eqn.

The Left Member of $(1)$ is obtained using the Distance Formula, whereas, the Right one uses the Formula for the Perp. Dist. of a Pt. to a Line.

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