Question

If two tangents to the parabola $y^2=4ax$ from a point P make angle ${\theta }_1$ and ${\theta }_2$ with the axis of the parabola, then the find the locus of P in each of the following cases.
${\theta }_1+{\theta }_2=\alpha$(a constant}
${\theta }_1+{\theta }_2=\frac{\pi }{2}$
$tan{\theta }_1+tan{\theta }_2=\lambda$ ( is constant)

Answer 1

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

Point $P:(h,k)$

Tangent to $y^2=4axisT:y=mx+\frac{a}{m}.......\left(ii\right)$

It passes through $(h,k)=>m^{2}h-mk+a=0.......\left(iii\right)($is a quadratic in$m)$

Let the roots to $\left(iii\right)$ be $m_1,m_2$

$=>m_{2=1}+m_2=\frac{k}{h};m_1{m}_2=\frac{a}{h};=>m_1-m_2=\frac{1}{h}\sqrt{k^2-4ah}......\left(v\right)$

$=>$locus of point of intersection$=>tan\beta =\pm \frac{m_1+m_2}{1-m_1{m}_2}=\frac{k}{h-a}........\left(iv\right)$

$=>locus=>y^2=\left({tan}^2\beta \right)(x+a{)}^2.......(vi)$

Given tangents make angle ${\theta }_1$and${\theta }_2$ with axis, $m_1=tan{\theta }_1$and$m_2=tan{\theta }_2$

Now $=>tan\beta =\pm \frac{tan{\theta }_1+{tan\theta }_2}{1-{tan\theta }_1{tan\theta }_2}=\pm tan({\theta }_1+{\theta }_2).......\left(vii\right)$

Putting values from $\left(vii\right)$ to $\left(vi\right)=>y^2={tan}^2({\theta }_1+{\theta }_2)({x-a)}^2.......(viii)$

Case I ${\theta }_1+{\theta }_2=\alpha ,$equation$\left(viii\right)$becomes$y^2={tan}^2\alpha ({x-a)}^2.....(I)$

Case II ${\theta }_1+{\theta }_2=\frac{\pi }{2}$equation$\left(viii\right)$becomes$y^2={tan}^2\frac{\pi }{2}({x+a)}^2$

$=>y^2=\frac{1}{0}({x-a)}^2$

$=>{(x-a)}^2=0$

$=>x=a($directrix$).........(II)$

Case III $tan{\theta }_1+{tan\theta }_2=\lambda$

Equation$\left(vii\right)=>y^2={\left(\frac{\lambda }{1-tan{\theta }_{1}tan{\theta }_2}\right)}^2({x-a)}^2$

$=>y^2(1-{tan{\theta }_{1}tan{\theta }_2)}^2={\lambda }^2({x-a)}^2($as$tan{\theta }_{1}tan{\theta }_2=m_1{m}_2=\frac{a}{h})$

$=>y^2(1-{\frac{a}{x})}^2={\lambda }^2({x-a)}^2$

$=>y^2={\lambda }^2{\left(x\right)}^2$

$=>y^2={\lambda }^2{x}^2............\left(III\right)$