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  • The equation of the parabola whose vertex is at (2,-1) and focus at (2,-3) is Option: 1 <img alt="\mathrm{ x^2+4 x-8 y-12=0}" src="/latex-image/

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

Option: 1

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


Option: 2

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


Option: 3

\mathrm{ x^2+8 y=12}


Option: 4

none of these


Answers (1)

best_answer

Vertex is (2,-1), Focus S is (2,-3)
Let the axis meet the directrix in Z. Let Z be (\alpha, \beta).
Since A is the mid-point of SZ,

\begin{array}{ll} \therefore & \mathrm{\frac{\alpha+2}{2}=2 \Rightarrow \alpha=2 ; \quad \frac{\beta-3}{2}=-1 \Rightarrow \beta=1 }\\ \therefore \quad & \text {Z is }(2,1) \end{array}
Slope of \mathrm{ZS =\frac{-3-1}{2-2}=\infty}

\begin{aligned} &\mathrm{ \therefore \quad Z \mathrm{~S} \text { is }|| \text { to } y \text {-axis. } }\\ &\mathrm{ \therefore \quad \text { directrix is }|| \text { to } x \text {-axis. } }\\ &\mathrm{ \therefore \quad \text { equation of the directrix is } y=1}\\ \end{aligned}

Now \mathrm{PS =\sqrt{(x-2)^2+(y+3)^2}}
PM = Length of the \perp from \mathrm{P} upon directrix \mathrm{=|y-1|}

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

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

Posted by

SANGALDEEP SINGH

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