Question

A line passing througjh the point of intersection of $x+y=4$ and $x-y=2$ makes an angle of ${\tan }^{-1}\frac{3}{4}$ with the $x$ axis. It intersects the parabola $y^2=4(x-3)$ at point $(x_1,y_1)$ 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 $\frac{32}{9}$

The point of intersection of the equations $x+y=4$ and
$x-y=2$ is $(3,1)$.
The equation of line through the point $(3,1)$ making an angle
${\tan }^{-1}\frac{3}{4}$ with the x-axis is calculated as
$y=\frac{3}{4}(x-3)+1y=\frac{3}{4}\left(x\right)-\frac{5}{4}$
Putting this in the equation of parabola $y^2=4(x-3)$ we have
${\left(\frac{3x-5}{4}\right)}^2=4(x-3)\Rightarrow 9x^2-94x+217=0x_1+x_2=\frac{94}{9},x_1{x}_2=\frac{217}{9}\Rightarrow |x_1-x_2|=\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}\Rightarrow |x_1-x_2|=\sqrt{{\left(\frac{94}{9}\right)}^2-4\left(\frac{217}{9}\right)}\Rightarrow |x_1-x_2|=\frac{32}{9}$