Question

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

Answer: (B)

Point of intersection of $x+y=4$ and $x-y=2is\equiv \left(3,1\right)$
The line though this making an angle $ta{n}^{-1}\frac{3}{4}$ with the x-axis
is $\left(y-1\right)=\frac{3}{4}\left(x-3\right)$
$\Rightarrow y=\frac{3x}{4}-\frac{5}{4}=\frac{3x-5}{4}$
Putting y in $y^2=4(x-3)$, we have
$9x^2-94x+217=0$
$\Rightarrow x_1+x_2=\frac{94}{9}$ and $x_1{x}_2=\frac{217}{9}$
$\Rightarrow \left|x_1-x_2\right|=\sqrt{{\left(x_1+x_2\right)}^2-4x_1{x}_2}=\frac{32}{9}$