Question

The line 2x-y+4=0 cuts the parabola $y^2=8x$ in P and Q.The mid-point of PQ is

• A. (1,2)
• B. (1,-2)
• C. (-1,2)
• D. (-1,-2)

Answer 1

The correct option is C

(-1,2)

Let the coordinates of P and Q be $(a{t}_1^2,2a{t}_1)$ and $(a{t}_2^2,2a{t}_2)$ respectively

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

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

Putting 4a=8

a=2

$\therefore \text{Focus of the given parabola=(a,0)=(2,0)}$

Using equation (2):

$\frac{4(t_2-t_1)}{2(t_2^2-t_1^2)}=2$

$\frac{(t_2-t_1)}{(t_2^2-t_1^2)}=1$

$\Rightarrow \frac{1}{t_2+t_1}=1$

$\Rightarrow t_1+t_2=1$

As,points P and Q lie on 2x-y+4=0

$P(a{t}_1^2,2a{t}_1)orP(2a{t}_1^2,4t_1)$ lie on line 2x-y+4=0

$\Rightarrow 2\left(2t_1^2\right)-\left(4t_1\right)+4=0$

$\Rightarrow t_1^2-t_1+1=0$ ...(3)

Also,$Q(a{t}_2^2,2a{t}_2)orP(2t_2^2,4t_2)$ lie on line 2x-y+4=0

$\Rightarrow 2\left(2t_2^2\right)-\left(4t_2\right)+4=0$

$\Rightarrow t_2^2-t_2+1=0$ ...(4)

Adding (3) and (4), we get,

$\Rightarrow t_1^2-t_1+1+t_2^2-t_2+1=0$

$\Rightarrow (t_1^2+t_2^2)-(t_1+t_2)+2=0$

$\Rightarrow (t_1^2+t_2^2)-1+2=0$

$[t_1+t_2=1,provedabove]$

$\Rightarrow (t_1^2+t_2^2)=-1$

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