Question

A line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $ta{n}^{-1}(3/4)$ 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 $(y-1)=\frac{3}{4}(x-3)$
$\Rightarrow 4y-4=3x-9$
$\Rightarrow 3x-4y=5$
Now for the intersection point of the line (i) with
parabola $y^2=4(x-3).$ Put $y=\left(\frac{3x-5}{4}\right)$,
then we get $\frac{(3x-5{)}^2}{16}=4(x-3)$
$\Rightarrow 9x^2+25-30x=64x-192$
$\Rightarrow 9x^2-94x+217=0$
$\Rightarrow x=\frac{94\pm \sqrt{8836-7812}}{18}=\frac{94\pm \sqrt{1024}}{18}$
$\Rightarrow x=\frac{94\pm 32}{18}=\frac{126}{18}$ or $\frac{62}{18}$
$\Rightarrow x_1=\frac{21}{3}=7$
and $x_2=\frac{31}{9}$
$\therefore \left|x_1-x_2\right|=\left|7-\frac{31}{9}\right|=\left|\frac{32}{9}\right|=\frac{32}{9}$