Let y^2=12 x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that angle OPA=90^°. Then the locus of the centroid of such triangles OPA is:

m_ AP=frac- t2 Equation of AP is y-6 t=(-t / 2)(x-3 t^2) Put y=0 => x=12+3 t^2 => A(12+3 t^2, 0) Let centroid of triangle OPA be G( h, k) => 3 h=0+3 t^2+12+3 t^2 3 k=0+6 t+0 => t=frac k2, h=2 t^2+4 => h=2 frac k^24+4 ⇒ Locus of ( h, k) is y^2=2 x-8

Let y 2 12 x be the parabola with its vertex at O Let P... | null (Mathematics)

\mathrm{m}_{\mathrm{AP}}=\frac{-\mathrm{t}}{2}\\

Equation of AP is

\\y-6 t=\frac{-t}{2}\left(x-3 t^{2}\right)\\

Put

\mathrm{y}=0 \Rightarrow \mathrm{x}=12+3 \mathrm{t}^{2}\\

\Rightarrow \mathrm{A}\left(12+3 \mathrm{t}^{2}, 0\right)\\

Let centroid of

\triangle

OPA be

\mathrm{G}(\mathrm{h}, \mathrm{k})\\

\Rightarrow 3 \mathrm{~h}=0+3 \mathrm{t}^{2}+12+3 \mathrm{t}^{2}\\

3 \mathrm{k}=0+6 \mathrm{t}+0\\

\Rightarrow \mathrm{t}=\frac{\mathrm{k}}{2}, \mathrm{~h}=2 \mathrm{t}^{2}+4\\

\Rightarrow \mathrm{h}=2 \frac{\mathrm{k}^{2}}{4}+4\\

⇒ Locus of

(\mathrm{h}, \mathrm{k})

\\y^{2}=2 x-8

Explanation

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