Question

Two tangents to a parabola $y^2=4ax$ meet at an angle of ${45}^0$. Prove that the locus of their point of intersection is the curve $y^2-4ax=(x+a{)}^2$.

Answer 1

Parabola: $y^2=4ax.......\left(i\right)$

Let the point of intersection be $(h,k)$

Tangent to $\left(i\right)$ be $T:y=mx+\frac{a}{m}.......\left(ii\right)$

$(h,k)$ lies on $T=>m^{2}h-mk+a=0........\left(iii\right)\left(quadraticinm\right)$

$=>m_1+m_2=\frac{k}{h},m_1{m}_1=\frac{a}{h},m_1-m_2=\frac{1}{h}\sqrt{k^2-4ah}$

So let the angle between two tangents $=\alpha$

$=>tan\alpha =\pm \frac{m_1-m_2}{1+m_1{m}_2}=\pm \frac{\sqrt{k^2-4ah}}{h+a}$

$=>k^2-4ah={tan}^2\alpha {(h+a)}^2$

Locus: $y^2-4ax={tan}^2\alpha ({x+a)}^2..........(iv)$

As given $\alpha ={45}^{\circ }=>tan\alpha =1$

So, equation $\left(iv\right)$ becomes $y^2-4ax={\left(1\right)}^2({x+a)}^2$

$=>y^2-4ax=({x+a)}^2$ is locus of point of intersection of tangents with angle between them equal to ${45}^{\circ }$.