Question

lf the equation $x^2+p{y}^2+y=a^2$ represents a pair of perpendicular lines, then the point of intersection of the lines is

• A. $(1,a)$
• B. $(1,-a)$
• C. $(0,a)$
• D. $(0,2a)$

Answer 1

The correct option is C $(0,a)$

$x^2+p{y}^2+y=a^2$ pair of perpendicular lines then
$coefficientof{x}^2+coefficientof{y}^2=0$
$1+p=0$
$\therefore p=-1$
$\therefore x^2-y^2+y=a^2$
So, for point of intersection of lines
Partial diff. w. r. f. $x:2x=0$ ----(1)
Partial diff. w. r. f. $y:-2y+1=0$ ----(2)
Solve 1 & 2
$x=0,y=\frac{1}{2}$
$(0,\frac{1}{2})$ is a point of intersection
$=0-(\frac{1}{2}{)}^2+\frac{1}{2}=a^2$
$\frac{1}{4}=a^2$
$\Rightarrow a=\frac{1}{2}$
$\therefore$ point of intersection is $(0,\frac{1}{2})$ $\Rightarrow (0,a)$