Question

A line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle ${\tan }^{-1}\left(\frac{3}{4}\right)$ 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

The correct option is B $\frac{32}{9}$

Given lines are $x+y=4$ 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 slope
$\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$ is $(y-1)=\frac{3}{4}(x-3)$
$\Rightarrow$ $4y-4=3x-9$
$\Rightarrow$ $3x-4y=\mathrm{5.....}\left(i\right)$
Now for the intersection point of the line $\left(i\right)$ with parabola $y^2=4(x-3)$. Put $y=\left(\frac{3x-5}{4}\right)$, then we get
$\frac{{(3-5x)}^2}{16}=4(x-3)$
$\Rightarrow 9x^2+25-30x=54x-192$
$\Rightarrow 9x^2-94x+217=0$
$\Rightarrow x=\frac{94\pm \sqrt{8836-7812}}{18}$
$=\frac{94\pm \sqrt{1024}}{18}$ $\Rightarrow x=\frac{94\pm 32}{18}=\frac{126}{8}or\frac{62}{18}$
$\Rightarrow x_1=\frac{21}{3}=7;x_2=\frac{31}{9}$
$|x_1-x_2|=|7-\frac{31}{9}|=\left|\frac{32}{9}\right|=\frac{32}{9}$