Question

The line 2x − y + 4 = 0 cuts the parabola y² = 8x in P and Q. The mid-point of PQ is
(a) (1, 2)
(b) (1, −2)
(c) (−1, 2)
(d) (−1, −2)

Answer 1

(c) (−1, 2)

Let the coordinates of P and Q be $\left(a{t_1}^2,2a{t}_1\right)$ and $\left(a{t_2}^2,2a{t}_2\right)$, respectively.

Slope of PQ = $\frac{2a{t}_2-2a{t}_1}{a{t_2}^2-a{t_1}^2}$ ......(1)

But, the slope of PQ is equal to the slope of 2x − y + 4 = 0.

∴ Slope of PQ = $\frac{-2}{-1}=2$

From (1),

$\frac{2a{t}_2-2a{t}_1}{a{t_2}^2-a{t_1}^2}=2$ .....(2)

Putting 4a = 8,
a = 2

∴ Focus of the given parabola = (a, 0) = $\left(2,0\right)$.

Using equation (2):

$\frac{4\left(t_2-t_1\right)}{2\left({t_2}^2-{t_1}^2\right)}=2$
$\frac{\left(t_2-t_1\right)}{\left({t_2}^2-{t_1}^2\right)}=1$
$\Rightarrow t_1+t_2=1$
As, points P and Q lie on 2x-y+4=0

$$\begin{gathered} \Rightarrow P(a{t_1}^2,2a{t}_1)\mathrm{or}P(2{t_1}^2,4t_1)\mathrm{lie}\mathrm{on}\mathrm{line}2x-y+4=0 \\ \Rightarrow 2\left(2{t_1}^2\right)-\left(4t_1\right)+4=0 \\ \Rightarrow {t_1}^2-t_1+1=0...\left(3\right) \\ Also,Q(a{t_2}^2,2a{t}_2)\mathrm{or}P(2{t_2}^2,4t_2)\mathrm{lie}\mathrm{on}\mathrm{line}2x-y+4=0 \\ \Rightarrow 2\left(2{t_2}^2\right)-\left(4t_2\right)+4=0 \\ \Rightarrow {t_2}^2-t_2+1=0...\left(4\right) \\ \mathrm{Adding}\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}, \\ \Rightarrow {t_1}^2-t_1+1+{t_2}^2-t_2+1=0 \\ \Rightarrow \left({t_1}^2+{t_2}^2\right)-\left(t_1+t_2\right)+2=0 \\ \Rightarrow \left({t_1}^2+{t_2}^2\right)-1+2=0\left[t_1+t_2=1,\mathrm{proved}\mathrm{above}\right] \\ \Rightarrow \left({t_1}^2+{t_2}^2\right)=-1 \end{gathered}$$

Let $\left(x_1,y_1\right)$ be the mid-point of PQ.

Then, we have:
$y_1=\frac{2a{t}_2+2a{t}_1}{2}=2\left(t_1+t_2\right)=2$
And, $x_1=\frac{a{t_1}^2+a{t_2}^2}{2}={t_1}^2+{t_2}^2=-1$

⇒ $\left(x_1,y_1\right)=\left(-1,2\right)$