Co-normal Points - Practice Questions & MCQ

Co-normal Points:

In general, a maximum of three normals can be drawn from a point to a parabola and their feet (points) where they meet the parabola are called the co-normal points.

P (h,k) is any point on the plane
equation of normal to the parabola $\mathrm{y}^2=4 \mathrm{ax}$ is

$
\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3
$
It passes through the point $\mathrm{P}(\mathrm{h}, \mathrm{k})$ then,

$
\begin{aligned}
& \mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3 \\
& \text { or, } \mathrm{am}^3+\mathrm{m}(2 \mathrm{a}-\mathrm{h})+\mathrm{k}=0
\end{aligned}
$
This cubic equation have three roots say, $m_1, m_2, m_3$

$
\begin{aligned}
& \mathrm{m}_1+\mathrm{m}_2+\mathrm{m}_3=0 \\
& \mathrm{~m}_1 \mathrm{~m}_2+\mathrm{m}_2 \mathrm{~m}_3+\mathrm{m}_3 \mathrm{~m}_1=\frac{2 \mathrm{a}-\mathrm{h}}{\mathrm{a}} \\
& \mathrm{~m}_1 \mathrm{~m}_2 \mathrm{~m}_3=-\frac{\mathrm{k}}{\mathrm{a}}
\end{aligned}
$

Points A, B, C in which the three normals from P (h,k) meets the parabola are called co-normal points 

Pair of Tangent of a Parabola

The combined equation of the pair of tangents drawn from an external point P(x₁,y₁) to the parabola, say S=y²-4ax=0 is  SS₁=T².

Where,

$\begin{aligned} & S=y^2-4 a x \\ & S_1=y_1^2-4 a x_1 \\ & T=y_1-2 a\left(x+x_1\right) \\ & \left(y^2-4 a x\right)\left(y_1^2-4 a x_1\right)=\left(y_1-2 a\left(x+x_1\right)\right)^2\end{aligned}$

Let (h, k) be any point on either of the tangents drawn from the point P (x₁, y₁). The equation of the line joining the point  (x₁, y₁)  and (h, k) is

$
\begin{aligned}
\mathrm{y}-\mathrm{y}_1 & =\frac{\mathrm{k}-\mathrm{y}_1}{\mathrm{~h}-\mathrm{x}_1}\left(\mathrm{x}-\mathrm{x}_1\right) \\
\Rightarrow \quad \mathrm{y} & =\frac{\mathrm{k}-\mathrm{y}_1}{\mathrm{~h}-\mathrm{x}_1} \mathrm{x}+\frac{\mathrm{hy}_1-\mathrm{kx}_1}{\mathrm{~h}-\mathrm{x}_1}
\end{aligned}
$
For this to be a tangent to the parabola it must be of the form $y=m x+\frac{a}{m}$ comparing this two equations, we get

$
m=\frac{k-y_1}{h-x_1} \text { and } \frac{a}{m}=\frac{h y_1-k x_1}{h-x_1}
$
Therefore, by multiplication we get

$
\begin{aligned}
& a=\left(\frac{k-y_1}{h-x_1}\right)\left(\frac{h y_1-k x_1}{h-x_1}\right) \\
\Rightarrow & a\left(h-x_1\right)^2=\left(k-y_1\right)\left(h y_1-k x_1\right)
\end{aligned}
$

Hence, the locus of the point (h, k) is

$\begin{aligned} & \Rightarrow \quad a\left(x-x_1\right)^2=\left(y-y_1\right)\left(x y_1-y x_1\right) \\ & \Rightarrow \quad\left(y^2-4 a x\right)\left(y_1^2-4 a x_1\right)=\left\{y y_1-2 a\left(x+x_1\right)\right\}^2 \\ & \Rightarrow \quad S S_1=T^2\end{aligned}$

Note:

The formula $\mathbf{S S}_1=\mathrm{T}^2$ works for finding pair of tangents to any general parabola as well.