Question

Tangents are drawn to a parabola at points whose abscissae are in the ratio $\mu :1;$ prove that they intersect on the curve $y^2=({\mu }^{\frac{1}{4}}+{\mu }^{-\frac{1}{4}}{)}^{2}ax.$

Answer 1

Let the point of contact of tangents be $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ and their point of intersection be $(h,k)$

Given $\frac{a{t}_2^2}{a{t}_1^2}=\frac{\mu }{1}$

$t_2^2=\mu t_1^2{t}_2={\mu }^{\frac{1}{2}}{t}_1.......\left(i\right)$

Point of intersection of tangents is $(a{t}_1{t}_2,a(t_1+t_2\left)\right)$

$\Rightarrow h=a{t}_1{t}_2.......\left(ii\right)$

Substituting $\left(i\right)$ in $\left(ii\right)$

$h=a{t}_1{(\mu }^{\frac{1}{2}}{t}_1)h=a{\mu }^{\frac{1}{2}}{t}_1^2\Rightarrow t_1^2=\frac{h}{a{\mu }^{\frac{1}{2}}}.........(iii)$

Also $k=a(t_1+t_2)$

Squaring both sides, we get

$k^2=a^2(t_1^2+t_2^2+2t_1{t}_2)$

Using $\left(i\right),\left(ii\right)$ and $\left(iii\right)$

$k^2=a^2(t_1^2+\mu t_1^2+2\frac{h}{a})k^2=a^2\left\{\right(\mu +1)t_1^2+2\frac{h}{a}\}{k}^2=a^2\left\{\right(\mu +1)\frac{h}{a\sqrt{\mu }}+2\frac{h}{a}\}{k}^2=ah(\sqrt{\mu }+\frac{1}{\sqrt{\mu }}+2)k^2=ah{({\mu }^{\frac{1}{4}}+{\mu }^{-\frac{1}{4}})}^2$

Replacing $h$ by $x$ and $k$ by $y$

$y^2={({\mu }^{\frac{1}{4}}+{\mu }^{-\frac{1}{4}})}^{2}ax$

Hence proved