Question

The product of the lengths of the perpendiculars drawn from the point $(−1,5)$ to the pair of lines $2x^2−xy−3y^2+6x+y+4=0$ 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
$2x^2−xy−3y^2+6x+y+4=0$
$⇒2x^2+(−y+6)x+\left(−3y^2+y+4\right)=0$
$∴x=\frac{(y−6)Â\pm \sqrt{(y−6{)}^6−4×2\left(−3y^2+y+4\right)}}{2×2}$
$=\frac{(y−6)Â\pm \sqrt{y^2+36−12y+24y^2−8y−32}}{4}$
$=\frac{(y−6)Â\pm \sqrt{25y^2+4−20y}}{4}$
$=\frac{(y−6)Â\pm \sqrt{(5y−2{)}^2}}{4}$
$x=\frac{(y−6)Â\pm (5y−2)}{4}$
So, $x=\frac{y−6+(5y−2)}{4}$ and $x=\frac{y−6−5y+2}{4}$
$⇒4x=6y−8\text{ and }4x=−4y−4$
$⇒2x−3y+4=0$ and $x+y+1=0$
Now, product length of perpendicular from $(−1,5)$ to both lines are
$P_1â‹\dots P_2=â\frac{−2−3(5)+4}{\sqrt{2^2+(−3{)}^2}}ââ\frac{−1+5+1}{\sqrt{1^2+1^2}}â$
$=\frac{13}{\sqrt{13}}×\frac{5}{\sqrt{2}}=\frac{65}{\sqrt{26}}$