Question

The locus of the point of intersection of tangents to the parabola $y^2=4ax$ which includes an angle $\alpha$ is

• A. $y^2-4ax=(x-a{)}^2{\tan }^2\alpha$
• B. $y^2-4ax=(x+a{)}^2{\cot }^2\alpha$
• C. $y^2-4ax=(x+a{)}^2{\tan }^2\alpha$
• D. $y^2-4ax=(x-a{)}^2{\cot }^2\alpha$

Answer 1

The correct option is C $y^2-4ax=(x+a{)}^2{\tan }^2\alpha$

The tangent to the parabola is
$y=mx+\frac{a}{m}$
Let the point of intersection be $T(h,k)$
It passes through the point $T(h,k)$, we have
$m^{2}h-mk+a=0\cdots \left(1\right)$
Let $m_1$ and $m_2$ be the roots of this equation, then
$m_1+m_2=\frac{k}{h}\cdots \left(2\right)m_1{m}_2=\frac{a}{h}\cdots \left(3\right)$

Then the equation of the tangents $TP$ and $TQ$ are
$y=m_{1}x+\frac{a}{m_1}\text{and}y=m_{2}x+\frac{a}{m_2}$

Hence, we have
$\tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\sqrt{(m_1+m_2{)}^2-4m_1{m}_2}}{1+m_1{m}_2}\right|$
Now from equations $\left(2\right)$ and $\left(3\right)$, we get
$\tan \alpha =\left|\frac{\sqrt{\frac{k^2}{h^2}-4\frac{a}{h}}}{1+\frac{a}{h}}\right|\Rightarrow \tan \alpha =\left|\frac{\sqrt{k^2-4ah}}{a+h}\right|\therefore k^2-4ah=(a+h{)}^2{\tan }^2\alpha$

Hence,the locus of point $T$ is
$y^2-4ax=(a+x{)}^2{\tan }^2\alpha$