The tangent \(P T\) and the normal \(P N\) to the parabola \(y^2=4 a x\) at a point \(P\) on it meet its…
- vertex is \(\left(\frac{2 a}{3}, 0\right)\)
- directrix is \(x=0\)
- latus rectum is \(\frac{2 a}{3}\)
- focus is \((a, 0)\)
Solution
\(y^2=4 a x\)
\(\therefore \quad\) Tangent to the parabola at \(P\) is \(y=\frac{x}{t}+a t\),
which meets the axis of parabola i.e \(x\)-axis \(a t\) \(T\left(-a t^2, 0\right)\).
Also normal to parabola at \(P\) is \(t x+y=2 a t+a t^3\)
which meets the axis of parabola at \(N\left(2 a+a t^2, 0\right)\)
Let \(G(x, y)\) be the centriod of \(\triangle P T N\), then
\(x=\frac{a t^2-a t^2+2 a+a t^2}{3} \text { and } y=\frac{2 a t}{3}\)
\(\Rightarrow \quad x=\frac{2 a+a t^2}{3} \quad \ldots\) (i) and \(y=\frac{2 a t}{3} \quad \ldots\) (ii)
Eliminating \(t\) from (i) and (ii), we get the locus of centriod \(G\) as \(3 x=2 a+a\left(\frac{3 y}{2 a}\right)^2 \Rightarrow y^2=\frac{4 a}{3}\left(x-\frac{2}{3} a\right)\),
which is a parabola with vertex \(\left(\frac{2 a}{3}, 0\right)\),
directrix as \(x-\frac{2 a}{3}=-\frac{a}{3} \Rightarrow x=\frac{a}{3}\),
latus rectum as \(\frac{4 a}{3}\) and focus as \((a, 0)\).
Asked in: JEE Advanced 2009 (Paper 2)