Question

Two manually perpendicular tangent of the parabola $y^2=4ax$ meet the axis in $P_1$ and $P_2$. If $S$ is the focus of the parabola, then $\frac{1}{\left(S{P}_1\right)}+\frac{1}{\left(S{P}_2\right)}$ is equal to :-

• A. $\frac{4}{a}$
• B. $\frac{2}{a}$
• C. $\frac{1}{a}$
• D. $\frac{1}{4a}$

Answer 1

The correct option is C $\frac{1}{a}$

$y^2=4ax............\left(1\right)$

Let the two mutually perpendicular tangents have slopes $m_1,m_1$ where $m_1=\frac{1}{m_2}$

And hence their questions be,

$T_1:y=mx+\frac{a}{m}...........\left(2\right)$

$T_2:y=\frac{1}{m}x-am...........\left(3\right)$

Let $P_1$and$P_2$be$(x_1,0)\&(x_2,0)$ these points must lie on $T_1$and$T_2$

For $T_1,o=m\left(x_1\right)+\frac{a}{m}$

$=>x_1=-\frac{a}{m^2}$

For $T_2,o=-\frac{x_2}{m}-am$

$=>x_2=-a{m}^2$

So, $P_1(\frac{-a}{m^2},0)$and$P_2(-a{m}^2,0)$

Focus $S=(a,o)$

Now $S{P}_1=\sqrt{(a+\frac{a}{m^2}{)}^2+0}=(a+\frac{a}{m^2})$

$S{P}_2=\sqrt{(a+a{m}^2{)}^2+0}=(a+a{m}^2)$

Now, $\frac{1}{{SP}_1}+\frac{1}{{SP}_2}=\frac{m^2}{a{m}^2+a}+\frac{1}{a+{am}^2}$

$=\frac{m^2+1}{a(m^2+1)}$

$=\frac{1}{a}$