Question

From an external point P tangents are drawn to the parabola; find the equation to the locus of P when these tangents make angles ${\theta }_1$ and ${\theta }_2$ with the axis, such that

$cos{\theta }_{1}cos{\theta }_2$ is constant $(=\mu )$.

Answer 1

Let the point P be $(a{t}_1{t}_2,a(t_1+t_2\left)\right)$

The tangent equations are $t_{1}y=x+a{t}_1^2$ and $t_{2}y=x+a{t}_2^2$

Since the tangents make angles ${\theta }_1$ and ${\theta }_2$ with the axis, $\frac{1}{t_1}=\tan {\theta }_1$ and $\frac{1}{t_2}=\tan {\theta }_2$

$\Rightarrow \cos {\theta }_1=\frac{t_1}{\sqrt{t_1^2+1}},\cos {\theta }_2=\frac{t_2}{\sqrt{t_2^2+1}}$

According to the given condition, $\frac{t_1}{\sqrt{t_1^2+1}}\times \frac{t_2}{\sqrt{t_2^2+1}}=\mu$

$\Rightarrow (t_1{t}_2{)}^2={\mu }^2\times [(t_1{t}_2{)}^2+t_1^2+t_2^2+1]$

$\Rightarrow (t_1{t}_2{)}^2={\mu }^2\times [(t_1{t}_2{)}^2+(t_1+t_2{)}^2-2t_1{t}_2+1]$

Let $a{t}_1{t}_2=x$ and $a(t_1+t_2)=y$

$\therefore {\left(\frac{x}{a}\right)}^2={\mu }^2\times [\frac{x^2}{a^2}+\frac{y^2}{a^2}-2\frac{x}{a}+1]$

$\Rightarrow x^2={\mu }^2(x^2+y^2-2ax+a^2)$

$\therefore x^2={\mu }^2(x-a{)}^2+{\mu }^2{y}^2$ is the required locus