Question

The equation $2x^2+4xy-p{y}^2+4x+qy+1=0$ will represent two mutually perpendicular straight lines, if

• A. $p=1$ and $q=2$ or $6$
• B. $p=-2$ and $q=2$ or $8$
• C. $p=2$ and $q=0$ or $8$
• D. $p=2$ and $q=0$ or $6$

Answer 1

The correct option is C $p=2$ and $q=0$ or $8$

Given pair

$2x^2+4xy-p{y}^2+4x+qy+1=0$

On comparing above eq with general form of equation of pair

$a=2,b=-p,h=2,g=2,f=\frac{q}{2},c=1$

As given lines are perpendicular

$a+b=0$

$2-p=0\Rightarrow p=2$

$2x^2+4xy-2y^2+4x+qy+1=0$

$y=\frac{-(4x+q)\pm \sqrt{16x^2+q^2+8qx+16x^2+32x+8}}{-4}$

$-4y=-(4x+q)\pm \sqrt{32x^2+8qx+32x+(8+q^2)}$

$-4y=-(4x+q)\pm \sqrt{(\sqrt{32}x{)}^2+8qx+32x+(\sqrt{(}8+q^2){)}^2}$

$\sqrt{(\sqrt{32}x{)}^2+8qx+32x+(\sqrt{(}8+q^2){)}^2}$ should be perfect square is equal to $(\sqrt{32}x+\sqrt{(8+q^2)}{)}^2$

Hence $8qx+32x=2\times \sqrt{32}x\times \sqrt{8+q^2}$

$4(q+4)=\sqrt{32}\times \sqrt{8+q^2}$

On squaring both sides

$16(q+4{)}^2=32(8+q^2)$

$(q^2+16+8q)=2(8+q^2)$

$q^2+16+8q=16+2q^2)$

$8q=q^2$

$(q-8)q=0$

$q=0,8$

Hence $p=2$ and $q=0,8$