Question

A line passing through the point of intersection of $x + y = 4$ and $x - y = 2$ makes an angle $tan^{-1}(3/4)$ with the x-axis. It intersects the parabola $y^2 = 4(x-3)$ at points $(x_1, y_1)$ and $(x_2, y_2)$ respectively. Then $|x_1-x_2|$ is equal to

• A. $\frac{16}{9}$
• B. $\frac{32}{9}$
• C. $\frac{40}{9}$
• D. $\frac{80}{9}$

Answer 1

Answer: (B)

Given lines are
$x+y=4 \text { and } x-y=2$
On solving these lines, we get
$x=3$ and $y=1$
Now, the equation of line which passes through the intersection point $(3,1)$ having alopo
$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$ is $(y-1)=\frac{3}{4}(x-3)$
$\Rightarrow 4 \,y-4=3 \,x-9$
$\Rightarrow 3\,x-4\,y=5$
Now for the intersection point of the line (i) with
parabola $y^{2}=4(x-3) .$ Put $y=\left(\frac{3 x-5}{4}\right)$,
then we get $\frac{(3 x-5)^{2}}{16}=4(x-3)$
$\Rightarrow 9 \,x^{2}+25-30 \,x=64 \,x-192$
$\Rightarrow 9\,x^{2}-94 \,x+217=0$
$\Rightarrow x=\frac{94 \pm \sqrt{8836-7812}}{18}=\frac{94 \pm \sqrt{1024}}{18}$
$\Rightarrow x=\frac{94 \pm 32}{18}=\frac{126}{18}$ or $\frac{62}{18}$
$\Rightarrow x_{1}=\frac{21}{3}=7$
and $x_{2}=\frac{31}{9}$
$\therefore \left|x_{1}-x_{2}\right|=\left|7-\frac{31}{9}\right|=\left|\frac{32}{9}\right|=\frac{32}{9}$