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The equation of the parabola whose vertex is at (2, -1) and focus at (2, -3) is

Option: 1

x^{2}+4x-8y-12=0
 


Option: 2

x^{2}-4x+8y+12=0


Option: 3

x^{2}+8y=12

 


Option: 4

None of these


Answers (1)

best_answer

Vertex is (2, –1), Focus S is (2, –3)

We know the focus already. To get the equation of parabola we should try to get the equation of directrix

Let the axis meet the directrix in Z\left ( \alpha ,\beta \right )

Since A is the mid-point of SZ,

\therefore \frac{\alpha +2}{2}=2 \Rightarrow \alpha =2

\therefore \frac{\beta -3}{2}=-1\Rightarrow \beta =1

\therefore Z\: is\: (2, 1)

ZS is the axis and we know directrix is perpendicular to the axis

Slope\: of \: ZS = \frac{-3+1}{2-2}=\infty

\therefore ZS \: is \left | \: \right |to \: y\: -axis.

\therefore directrix\: \: is \left | \: \right |to \: x\: -axis.

Also directrix passes through (2,1), so the equation of directrix is y = 1

Appying PS = PM

\therefore \left ( x-2 \right )^{2}+\left ( y+3 \right )^{2}= \left ( y-1 \right )^{2}\\*\Rightarrow x^{2}-4x+4+y^{2}+6y+9=y^{2}-2y+1

\Rightarrow x^{2}-4x+8y+12=0

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