Question

Two tangents are drawn from the point $(-2,-1)$ to the parabola $y^2=4x.$ If $\alpha$ is the angle between these tangents, then the value of $\tan \alpha$ is

Answer 1

Given parabola is $y^2=4x.$
On comparing this with the standard equation of the parabola $y^2=4ax$
$\Rightarrow a=1.$
Any tangent having slope $m$ is$y=mx+\frac{a}{m}$
So here tangent is $y=mx+\frac{1}{m}$
It passes through $(-2,-1)$

On putting the point $(x_1,y_1)$ in the expression $y_1^2-4a{x}_1$, we can find out if the point is outside, inside or on the parabola.

Here, $(-1{)}^2-4(1\left)\right(-2)=1+8=9>0$
Since this is greater than zero, we know that the point lies outside the parabola.

$\Rightarrow 2m^2-m-1=0$
On solving this quadratic equation, let the values that we get are $m_1\text{and}{m}_2.$
$\Rightarrow m_1=1,m_2-\frac{1}{2}$

$\tan \alpha =|\frac{m_1-m_2}{1+m_1{m}_2}|$

$\Rightarrow \tan \alpha =|\frac{1+\frac{1}{2}}{1-\frac{1}{2}}|=3$