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Q.5. If the vertex and focus of a parabola are (3, 3) and (- 3, 3) respectively, then its equation is
(a) $x^2-6x+24y-63=0$
(b) $x^2-6x+24y+81=0$
(c) $y^2-6y+24x-63=0$
(d) $y^2-6y-24x+81=0$

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Vertex} \mathrm{and} \mathrm{focus} \mathrm{are} \left(3,3\right) \mathrm{and} \left(-3,3\right) \mathrm{which} \mathrm{lie} \mathrm{on} \mathrm{line} \mathrm{y}=3 \\ \mathrm{Hence} \mathrm{axis} \mathrm{of} \mathrm{parabola} \mathrm{is} \mathrm{horizontal} \mathrm{and} \mathrm{vertex} \mathrm{lies} \mathrm{rightward} \mathrm{of} \mathrm{cous} \mathrm{so} \mathrm{equation} \mathrm{will} \mathrm{be} \\ {\left(\mathrm{y}-\mathrm{k}\right)}^2=-4\mathrm{a}\left(\mathrm{x}-\mathrm{h}\right) \\ \mathrm{Where} \left(\mathrm{h},\mathrm{k}\right) \mathrm{are} \mathrm{vertex} \\ \mathrm{a}=\mathrm{Distance} \mathrm{between} \mathrm{vertex} \mathrm{and} \mathrm{focus} \\ =\sqrt{{\left(3+3\right)}^2+{\left(3-3\right)}^2}=6 \\ \mathrm{Hence} \mathrm{equation} \mathrm{is} \\ {\left(\mathrm{y}-3\right)}^2=-24\left(\mathrm{x}-3\right) \\ \Rightarrow {\mathrm{y}}^2+9-6\mathrm{y}=-24\mathrm{x}+72 \\ \Rightarrow {\mathrm{y}}^2-6\mathrm{y}+24\mathrm{x}-63=0 \end{gathered}$$

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