Question

Chord joining two distinct points $P({\alpha }^2,k_1)$ and $Q(k_2,-\frac{16}{\alpha })$ on the parabola $y^2=16x$ always passes through a fixed point. Find the co-ordinates of fixed point.

Answer 1

Given, $y^2=16x\Rightarrow y^2=4\left(4x\right)$

Comparing with general equation of a parabola, $y^2=4\left(ax\right)$, we get $a=4$.

Now, putting point P in the equation, ${k_1}^2=16{\alpha }^2\Rightarrow k_1=\pm 4\alpha$

Similarly, putting point Q in the equation,

${\frac{-16}{\alpha }}^2=16k_2\Rightarrow k_2=\frac{16}{{\alpha }^2}$

Now, the EQUATION OF CHORD:

Case-1: $k_1=4\alpha$,

We have, $y-4\alpha =\left(\frac{4\alpha +\frac{16}{\alpha }}{{\alpha }^2-\frac{16}{{\alpha }^2}}\right)(x-{\alpha }^2)\Rightarrow y-4\alpha =\frac{4(\alpha +\frac{4}{\alpha })(x-{\alpha }^2)}{(\alpha -\frac{4}{\alpha })(\alpha +\frac{4}{\alpha })}\Rightarrow (y-4\alpha )(\alpha -\frac{4}{\alpha })=4(x-{\alpha }^2)\Rightarrow \alpha y-4{\alpha }^2-\frac{4}{\alpha }y+16=4x-4{\alpha }^2\Rightarrow {\alpha }^{2}y-4y+16\alpha =4\alpha x\Rightarrow y({\alpha }^2-4)+4\alpha (4-x)=0$

So, in this case the point is (4,0).

Case-2: $k_1=-4\alpha$,

We have, $y+4\alpha =\left(\frac{-4\alpha +\frac{16}{\alpha }}{{\alpha }^2-\frac{16}{{\alpha }^2}}\right)(x-{\alpha }^2)\Rightarrow y+4\alpha =\frac{-4(\alpha -\frac{4}{\alpha })(x-{\alpha }^2)}{(\alpha -\frac{4}{\alpha })(\alpha +\frac{4}{\alpha })}\Rightarrow (y+4\alpha )(\alpha +\frac{4}{\alpha })=-4(x-{\alpha }^2)\Rightarrow \alpha y+4{\alpha }^2+\frac{4}{\alpha }y+16=-4x+4{\alpha }^2\Rightarrow {\alpha }^{2}y+4y+16\alpha =-4\alpha x\Rightarrow y({\alpha }^2+4)+4\alpha (4+x)=0$

So, in this case the point is (-4,0) which is not possible as it lies outside the curve.

Hence, the correct answer is (4,0)