The line $x-2 y-3=0$ cuts the parabola $y^2=4 \mathrm{ax}$ at the points P and Q . If the focus of this…
The line $x-2 y-3=0$ cuts the parabola $y^2=4 \mathrm{ax}$ at the points P and Q . If the focus of this parabola is $\left(\frac{1}{4}, k\right)$, then $P Q=$
$16 a \sqrt{5}$
$8 a \sqrt{5}$
$4 a \sqrt{5}$
$2 a \sqrt{5}$
Solution
Given the equation of the parabola $y^2=4 a x$ and focus of this parabola is $\left(\frac{1}{4}, k\right)$. $\Rightarrow(a, 0)=\left(\frac{1}{4}, k\right) \Rightarrow k=0, a=\frac{1}{4}$
Since, $x-2 y-3=0 \Rightarrow x=2 y+3$
Now, $y^2=4 \times \frac{1}{4} x \Rightarrow y^2=2 y+3$
$\begin{aligned}& \Rightarrow y^2-2 y+3=0 \Rightarrow(y-3)(y+1)=0 \\& \Rightarrow y=-1,3 \Rightarrow x=1,9\end{aligned}$
Let $\mathrm{P}(1,-1)$ and $\mathrm{Q}(9,3)$
$\begin{aligned} & P Q=\sqrt{8^2+4^2}=\sqrt{80}=4 \sqrt{5} \\ & =4 \times 4 \times \frac{1}{4} \sqrt{5}=16 a \sqrt{5}\end{aligned}$