Question

If the normals at three points P,Q and R of the parabola $y^2=4ax$ meet in a point O and S is its focus, then |SP|. |SQ|. |SR| is equal to

• A. $a^2$
• B. $a(SO{)}^3$
• C. $a(SO{)}^2$
• D. None of these

Answer 1

Answer: (C)

$a(SO{)}^2$

Given,

$P,Q,R$ are points on the parabola $y^2=4ax$ such that the normals drawn to them meet at point $O$.

$S$ is the focus of the parabola .

Now, the parametric equation of the normal drawn to a point on the parabola is

$y=mx-2am-a{m}^3$ where ${}^{\prime }{m}^{\prime }$ is the parameter, basically its the slope $-\left(i\right)$

equation $\left(i\right)$ is the normal at the point $({am}^2,-2am)$ of the parabola .

Say the normal in equation $\left(i\right)$ passes through a point $(\alpha ,\beta )$

$\therefore \beta =m\alpha -2am-a{m}^3-\left(ii\right)\Rightarrow a{m}^3+m(20-\alpha )+\beta =0-\left(iii\right)$

Now the three points can be parametrically represented as

$P\equiv (a{m}_1^2,-2a{m}_1)Q\equiv (a{m}_2^2,-2a{m}_2)R=(a{m}_3^2,-2a{m}_3)S=(a,0)$

So,

$SP=\sqrt{{(a-a{m}_1^2)}^2+{\left(2a{m}_1\right)}^2}=\sqrt{a^2+a^2{m}_1^4-2a^2{m}_1^2+4a^2{m}_1^2}=\sqrt{a^2+a^2{m}_1^4+2a^2{m}_1^2}=\sqrt{{(a+a{m}_1^2)}^2}=a(1+m_1^2)$

Similarly $SQ=a(1+m_2^2)\&SR=a(1+m_3^2)$

$\therefore \left|SP\right|.\left|SQ\right|.\left|SR\right|=a^3(1+m_1^2)(1+m_2^2)(1+m_3^2)=a^3(1+m_1^2+m_2^2+m_3^2+m_1^2{m}_2^2+m_1^2{m}_3^2+m_2^2{m}_3^2+m_1^2{m}_2^2{m}_3^2)$

$m_1,m_2,m_3$ are basically the roots of the equation $\left(ii\right),\left(iii\right)$

$m_1+m_2+m_3=0m_1{m}_2+m_2{m}_3+m_3{m}_1=\frac{2a-\alpha }{a}-\left(iv\right)m_1{m}_2{m}_3=\frac{\beta }{a}$

Using $\left(iv\right)$ we can simplify the expression for

$\left|SP\right|\left|SQ\right|\left|SR\right|=a[{(a-\alpha )}^2+{\beta }^2]=a{SO}^2$