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-ordinate of fixed point.

Answer 1

Slope of $PQ=\frac{y_2-y_1}{x_2-x_1}=\frac{k_1-\left(\frac{-16}{\alpha }\right)}{{\alpha }^2-k_2}$

$=\frac{k_1+\frac{16}{\alpha }}{{\alpha }^2-k_2}=\frac{16+k_1\alpha }{\alpha ({\alpha }^2-k_2)}$

Now we have $({\alpha }^2,k_1)$ passes through

$y^2=16x$

$\Rightarrow {k_1}^2=16{\alpha }^2\Rightarrow k_1=\pm 4\alpha$

and $(k_2,\frac{-16}{\alpha })$ passes through $y^2=16x$

$\Rightarrow {\left(\frac{-16}{\alpha }\right)}^2=\frac{256}{{\alpha }^2}=16k_2$ (or) $\frac{16}{{\alpha }^2}=k_2$

Equation of $PQ$ is $\frac{y-k_1}{x-{\alpha }^2}=$

slope$\Rightarrow \frac{y-k_1}{x-{\alpha }^2}=\frac{16+k_1\alpha }{\alpha ({\alpha }^2-k_2)}$

$=\frac{16+k_1\alpha }{\alpha ({\alpha }^2-\frac{16}{{\alpha }^2})}$ Since $k_2=\frac{16}{{\alpha }^2}$

$\Rightarrow \frac{y-k_1}{x-{\alpha }^2}=\frac{\alpha (16+k_1\alpha )}{{\alpha }^4-16}$

On cross multiplication , we get

$\Rightarrow (y-k_1)({\alpha }^4-16)=\alpha (x-{\alpha }^2)(16+k_1\alpha )$

$\Rightarrow y{\alpha }^4-16y-k_1{\alpha }^4+16k_1=(x-{\alpha }^2)(16\alpha +k_1{\alpha }^2)$

$\Rightarrow y{\alpha }^4-16y-k_1{\alpha }^4+16k_1=16x\alpha +k_{1}x{\alpha }^2-16{\alpha }^3-k_1{\alpha }^4$ on cancelling same terms on both sides

$\Rightarrow y{\alpha }^4-16y+16k_1=16x\alpha +k_{1}x{\alpha }^2-16{\alpha }^3$

$\Rightarrow y({\alpha }^4-16)-x(16\alpha +k_1{\alpha }^2)+(16k_1+16{\alpha }^3)=0$ is the required equation

we have $k_1=4\alpha$

$\Rightarrow y({\alpha }^4-16)-x(16\alpha +4{\alpha }^3)+(64\alpha +16{\alpha }^3)=0$

$\Rightarrow y({\alpha }^4-16)-4x(4\alpha +{\alpha }^3)+16(4\alpha +16{\alpha }^3)=0$

$\Rightarrow y({\alpha }^4-16)+(4\alpha +{\alpha }^3)(16-4x)=0$

This equation passes through

$y=0,x=4$ when $y=0,x=4$

$\therefore$ the fixed point is $(4,0)$