Question

The locus of a point which divides the line segment joining the point $(0,-1)$ and a point on the parabola, $x^2=4y$, internally in the ratio $1:2$, is

• A. $9x^2-12y=8$
• B. $4x^2-3y=2$
• C. $x^2-3y=2$
• D. $9x^2-3y=2$

Answer 1

The correct option is A

$9x^2-12y=8$

Explanation for the correct option:

Let the point on the parabola be $\left(2t,t^2\right)$ and the locus point be $\left(h,k\right)$

Section formula for $(x_1,y_1),(x,y)$and $(x_2,y_2)$ having ratio $m_1:m_2$ is

$\mathbf{x}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{1}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{y}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{y}}_{\mathbf{1}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}$

Using the section formula for $(0,-1)$ , $\left(h,k\right)$ and $\left(2t,t^2\right)$ having ratio $1:2$

$h=\frac{2\left(0\right)+1\left(2t\right)}{1+2},k=\frac{2(-1)+1\left(t^2\right)}{1+2}$

$h=\frac{2t}{3},k=\frac{t^2-2}{3}$

$$\begin{gathered} \Rightarrow 3h=2t,3k=t^2-2 \\ \Rightarrow t=\frac{3h}{2},3k+2=t^2 \end{gathered}$$

Substituting the value of $t$, we get

$$\begin{gathered} 3k+2={\left(\frac{3h}{2}\right)}^2 \\ 3k+2=\frac{9h^2}{4} \\ 12k+8=9h^2 \\ 9h^2-12k=8 \end{gathered}$$

Substituting $h=x,k=y$

$\Rightarrow 9y^2-12x=8$

Hence, option A is correct.