Question

Locus of trisection point of any arbitrary double ordinate of the parabola $x^2=4by$ is

• A. $9x^2=4by$
• B. $3x^2=2by$
• C. $9x^2=by$
• D. $9x^2=2by$

Answer 1

The correct option is A

$9x^2=4by$

Explanation for the correct options:

Find the locus:

Given, $x^2=4by$

Double ordinate of the parabola is perpendicular to the axis of the parabola.

Let $AB$ be the double ordinate intersecting at $A$ and $B$.

Using parametric form

$$\begin{gathered} A=\left(-2bt,b{t}^2\right) \\ B=\left(2bt,b{t}^2\right) \end{gathered}$$

By section formula,

$\mathbf{x}\mathbf{=}\frac{{\mathbf{mx}}_{\mathbf{1}}\mathbf{+}{\mathbf{nx}}_{\mathbf{2}}}{\mathbf{m}\mathbf{+}\mathbf{n}}\mathbf{,}\mathbf{y}\mathbf{=}\frac{{\mathbf{my}}_{\mathbf{1}}\mathbf{+}{\mathbf{ny}}_{\mathbf{2}}}{\mathbf{m}\mathbf{+}\mathbf{n}}$

Let $P\left(h,k\right)$ be the trisection point.

Ratio is $2:1=m:n$ and $$\begin{gathered} A=\left(-2bt,b{t}^2\right)=\left(x_2,y_2\right) \\ B=\left(2bt,b{t}^2\right)=\left(x_1,y_1\right) \end{gathered}$$

So,

$$\begin{gathered} h=\frac{2\times 2bt+1\times \left(-2bt\right)}{2+1} \\ =\frac{2bt}{3} \\ \Rightarrow t=\frac{3h}{2b} \end{gathered}$$ and $$\begin{gathered} k=\frac{2\times b{t}^2+1\times \left(b{t}^2\right)}{2+1} \\ =\frac{3b{t}^2}{3} \\ k=b{t}^2 \end{gathered}$$

So

$$\begin{gathered} \Rightarrow k=b{\left(\frac{3h}{2b}\right)}^2 \\ \Rightarrow k=\frac{9h^2}{4b} \\ \Rightarrow 4kb=9h^2 \end{gathered}$$

Put $\left(x,y\right)$ for $\left(h,k\right)$

$\Rightarrow 4by=9x^2$

Hence, option (A) is the correct answer