- A$(1, 2)$
- B$(1, -2)$
- ✓$(-1, 2)$
- D$(-1, -2)$
Let the coordinates of $P$ and $Q$ be $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2),$ respectively.
$\text{Slope of PQ }=\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}...(1)$
But, the slope of PQ is equal to the slope of $2x - y + 4 = 0.$
$\therefore\ \text{Slope of PQ}=\frac{-2}{-1}=2$
From $(1),$
$\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}=2\ ...(2)$
Putting $4a = 8,$
$a = 2$
$\therefore$ Focus of the given parabola $= (a, 0) = (2, 0)$
Using equation $(2):$
$\frac{4(\text{t}_2-\text{t}_1)}{2(\text{t}_2^2-\text{t}_1^2)}=2$
$\frac{(\text{t}_2-\text{t}_2)}{(\text{t}_2^2-\text{t}_1^2)}=1$
$\Rightarrow\ \text{t}_1+\text{t}_2=1$
As, points $P$ and $Q$ lie on $2x - y + 4 = 0$
$\Rightarrow P(at_1^2, 2at_1)$ or $P(2t_1^2, 4t_1)$ lie on line $2x - y + 4 = 0$
$\Rightarrow 2(2t_1^2) - (4t_1) + 4 = 0$
$\Rightarrow t_1^2 - t_1 + 1 = 0 ...(3)$
Also, $Q(at_2^2, 2at_2)$ or $P(2t_2^2, 4t_2)$ lie on line $2x - y + 4 = 0$
$\Rightarrow 2(2t_2^2) - (4t_2) + 4 = 0$
$\Rightarrow t_2^2 - t_2 + 1 = 0 ...(4)$
Adding $(3)$ and $(4),$ we get,
$\Rightarrow t_1^2 - t_1 + 1 + t_2^2 - t_2 + 1 = 0$
$\Rightarrow (t_1^2 + t_2^2) - (t_1 + t_2) + 2 = 0$
$\Rightarrow (t_1^2 + t_2^2) - 1 + 2 = 0 [t_1 + t_2 = 1,$ proved above$]$
$\Rightarrow (t_1^2 + t_2^2) = -1$
Let$ (x_1, y_1)$ be the mid-point of $PQ.$
Then, we have:
$\text{y}_1=\frac{2\text{at}_2+2\text{at}_1}{2}=2(\text{t}_1+\text{t}_2)=2$
And, $\text{x}_1=\frac{\text{at}_1^2+\text{at}_2^2}{2}=\text{t}_1^2+\text{t}_2^2=-1$
$\Rightarrow\ (\text{x}_1, \text{y}_1)=(-1, 2)$
