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. Itintersects 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. $\frac{1}{9}$
• B. $\frac{32}{9}$
• C. $\frac{40}{9}$
• D. $\frac{80}{9}$

Answer 1

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

The line $x+y=4$ and $x-y=2$ intersect at $(3,1)$

Equation of line through $(3,1)$ and having slope $\frac{3}{4}$ is

$(y-1)=\frac{3}{4}(x-3)$

$⟹y=\frac{3x-5}{4}$

$y^2=4(x-3)$

$\therefore {\left(\frac{3x-5}{4}\right)}^2=4(x-3)$

$\therefore 9x^2-94x+217=0$

$|x_1-x_2|=\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}$

$=\sqrt{(\frac{94}{9}{)}^2-4(\frac{217}{9})}$

$=\sqrt{\frac{1024}{81}}$

$=\frac{32}{9}$

So, the answer is option (B).