Question

The product of the lengths of the perpendiculars drawn from the point $(-1,5)$ to the pair of lines $2 x^{2}-x y-3 y^{2}+6 x+y+4=0$ is

• A. $\frac{68}{\sqrt{2}}$
• B. $\frac{68}{\sqrt{26}}$
• C. $\frac{65}{\sqrt{2}}$
• D. $\frac{65}{\sqrt{26}}$

Answer 1

Answer: (D)

The pair of straight line
$ 2 x^{2}-x y-3 y^{2}+6 x+y+4=0 $
$\Rightarrow 2 x^{2}+(-y+6) x+\left(-3 y^{2}+y+4\right)=0 $
$ \therefore x =\frac{(y-6) \pm \sqrt{(y-6)^{6}-4 \times 2\left(-3 y^{2}+y+4\right)}}{2 \times 2} $
$=\frac{(y-6) \pm \sqrt{y^{2}+36-12 y+24 y^{2}-8 y-32}}{4} $
$=\frac{(y-6) \pm \sqrt{25 y^{2}+4-20 y}}{4} $
$=\frac{(y-6) \pm \sqrt{(5 y-2)^{2}}}{4} $
$x=\frac{(y-6) \pm(5 y-2)}{4} $
So, $x = \frac{y - 6 + (5y - 2)}{4}$ and $x = \frac{y - 6 - 5y + 2}{4}$
$ \Rightarrow 4 x=6 y-8 \text { and } 4 x=-4 y-4 $
$\Rightarrow 2 x-3 y+4=0$ and $x+y+1=0$
Now, product length of perpendicular from $(-1,5)$ to both lines are
$P_{1} \cdot P_{2}=\left|\frac{-2-3(5)+4}{\sqrt{2^{2}+(-3)^{2}}}\right|\left|\frac{-1+5+1}{\sqrt{1^{2}+1^{2}}}\right|$
$ =\frac{13}{\sqrt{13}} \times \frac{5}{\sqrt{2}}=\frac{65}{\sqrt{26}}$