Question

If two tangents drawn from the point $P$ to the parabola $y^2=12x$ be such that the slope of the tangent is double the other, then ${}^{\prime }{P}^{\prime }$ lies on the curve

• A. $2y^2=9x$
• B. $y^2=9x$
• C. $2y^2=27x$
• D. $y^2=15x$

Answer 1

The correct option is C $2y^2=27x$

Given a parabola ,$y^2=12x$

There is a point $P$ outside the parabola with coordinates$(h,k)$.Now the equation of a tangent to the parabola $y^2=4ax$ is given by

$y=mx+\frac{a}{m};m$ is the slope of tangent

Here $a=3$

$y=mx+\frac{3}{m}$

is the equation of any tangent to the parabola $y^2=12x$.This tangent passes through point $(h,k)$ hence

$\Rightarrow m^{2}h-km+3=0-\left(i\right)$

So we can see that there are two possible values of ${}^{\prime }{m}^{\prime }$ and hence we can have two tangents drawn to the parabola $y^2=12x$ from the point $(h,k)$.

Also the slope of one is twice that of the other.

So say $m_1$ & $m_2$ are the roots of the quadratic equation$\left(i\right)$

Then we have

$m_1=2m_2-\left(ii\right)m_1+m_2=\frac{k}{h}-\left(iii\right)m_1.m_2=\frac{3}{h}-\left(iv\right)$

Now, $2m_2^2=\frac{3}{h}-\left(v\right)3m_2=\frac{k}{h}\Rightarrow m_2=\frac{k}{3h}-\left(vi\right)$

Using $\left(vi\right)$ we can reduce $\left(v\right)$ as $\frac{2k^2}{9h^2}=\frac{3}{h}$

$\Rightarrow 2k^2=27h$

Replace $k$ with $y$ & $h$ with $x$ we have

$2y^2=27x$