Search any question & find its solution
Question:
Answered & Verified by Expert
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 $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ respectively. Then, $\left|x_{1}-x_{2}\right|$ is equal to
Options:
Solution:
2015 Upvotes
Verified Answer
The correct answer is:
$\frac{32}{9}$
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 slope
$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$ is $(y-1)=\frac{3}{4}(x-3)$
$\Rightarrow \quad 4 y-4=3 x-9$
$\Rightarrow \quad 3 x-4 y=5$
Now for the intersection point of the line (i) with parabola $y^{2}=4(x-3)$. Put $y=\left(\frac{3 x-5}{4}\right)$
then we get $\quad \frac{(3 x-5)^{2}}{16}=4(x-3)$
$\Rightarrow \quad 9 x^{2}+25-30 x=64 x-192$
$\Rightarrow \quad 9 x^{2}-94 x+217=0$
$\Rightarrow \quad x=\frac{94 \pm \sqrt{8836-7812}}{18}=\frac{94 \pm \sqrt{1024}}{18}$
$\Rightarrow \quad x=\frac{94 \pm 32}{18}=\frac{126}{18}$ or $\frac{62}{18}$
$\Rightarrow \quad x_{1}=\frac{21}{3}=7$
and $\quad x_{2}=\frac{31}{9}$
$\therefore \quad\left|x_{1}-x_{2}\right|=\left|7-\frac{31}{9}\right|=\left|\frac{32}{9}\right|=\frac{32}{9}$
$$
x+y=4 \text { and } x-y=2
$$
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 slope
$\theta=\tan ^{-1}\left(\frac{3}{4}\right)$ is $(y-1)=\frac{3}{4}(x-3)$
$\Rightarrow \quad 4 y-4=3 x-9$
$\Rightarrow \quad 3 x-4 y=5$
Now for the intersection point of the line (i) with parabola $y^{2}=4(x-3)$. Put $y=\left(\frac{3 x-5}{4}\right)$
then we get $\quad \frac{(3 x-5)^{2}}{16}=4(x-3)$
$\Rightarrow \quad 9 x^{2}+25-30 x=64 x-192$
$\Rightarrow \quad 9 x^{2}-94 x+217=0$
$\Rightarrow \quad x=\frac{94 \pm \sqrt{8836-7812}}{18}=\frac{94 \pm \sqrt{1024}}{18}$
$\Rightarrow \quad x=\frac{94 \pm 32}{18}=\frac{126}{18}$ or $\frac{62}{18}$
$\Rightarrow \quad x_{1}=\frac{21}{3}=7$
and $\quad x_{2}=\frac{31}{9}$
$\therefore \quad\left|x_{1}-x_{2}\right|=\left|7-\frac{31}{9}\right|=\left|\frac{32}{9}\right|=\frac{32}{9}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.