$A(-2,3)$ is a fixed point outside the parabola $y^2=4 a x(a>0)$ and $P$ is a point moving on the parabola. The locus of point $Q$ which divides $A P$ in the ratio $3: 2$ is a conic. Then focus of that conic is

• A

  $(a, 0)$
• B

  $\left(\frac{-4}{5}+\frac{3 a}{5}, \frac{a}{5}\right)$
• Correct answer
  C

  $\left(\frac{3 a-4}{5}, \frac{6}{5}\right)$
• D

  $\left(\frac{a}{5}, \frac{3 a-4}{5}\right)$

Solution

Let $P$ be $\left(a t^2, 2 a t\right)$ and $Q$ be $(h, k)$. Also given, $A=(-2,3)$ Now, since $Q$ divides $A P$ in the ratio $3: 2$ i.e., $$ \frac{A Q}{Q P}=\frac{3}{2} $$
$$ \begin{array}{rlrl} \Rightarrow & \quad 5 k & =6 a t+6 \\ \Rightarrow \quad t & =\frac{5 k-6}{6 a} \text { put in Eq. (i), we get } \\ 5 h & =3 a\left(\frac{5 k-6}{6 a}\right)^2-4 \\ \Rightarrow \quad 5 h+4 & =\frac{(5 k-6)^2}{12 a} \Rightarrow(5 k-6)^2=60 a h+48 a \\ \Rightarrow \quad(5 k-6)^2 & =12 a(5 h+4) \end{array} $$ Hence, the locus of $Q(h, k)$ is
On comparing Eq. (iii) by $Y^2=4 A X$, we get, $$ \begin{aligned} \quad & Y & =5 y-6, X=5 x+4 \text { and } 4 A=12 a \\ & A & =3 a \end{aligned} $$ So, focus of $Y^2=4 A X$ is $(A, 0)$ i.e. $X=A$ and $Y=0$ Hence, focus of Eq. (iii) is $$ \begin{aligned} & & 5 x+4 & =3 a \text { and } 5 y-6=0 \\ \Rightarrow & & x & =\frac{3 a-4}{5} \text { and } y=\frac{6}{5} \\ & \text { i.e. } & (x, y) & =\left(\frac{3 a-4}{5}, \frac{6}{5}\right)=\text { Focus } \end{aligned} $$

Hence, the correct answer is option: 'C'.