Question

If the roots of the equation $(x_1^2-{\text{a}}^2){\text{m}}^2-2x_1{y}_1\text{m}+(y_1^2+{\text{b}}^2)=0,(\text{a}>\text{b)}$ are the slopes
of two perpendicular lines intersecting at $\text{P(}{\text{x}}_1{\text{y}}_1).$ then the locus of P is

• A. $x^2+y^2={\text{a}}^2\text{+}{\text{b}}^2$
• B. ${\text{x}}^2+{\text{y}}^2=a^2-b^2$
• C. ${\text{x}}^2-{\text{y}}^2={\text{a}}^2+{\text{b}}^2$
• D. ${\text{x}}^2-{\text{y}}^2={\text{a}}^2{\text{b}}^2$

Answer 1

The correct option is B ${\text{x}}^2+{\text{y}}^2=a^2-b^2$

Given eqn

$(x_1^2-a^2)m^2-2x_1{y}_{1}m+(y_1^2+b^2)=0.....\left(1\right)$

Let $A{m}^2+Bm+C$ be general eqn $.....\left(2\right)$

Comparing $\left(1\right)$ & $\left(2\right)$

$A(x_1^2-a^2)$ $B=-2x_1{y}_1$ $C=y_1^2+b^2$

Let $m_1$ & $m_2$ are roots of the given equation

We know $m_1+m_2=\frac{-B}{A}$ $m_1{m}_2=\frac{C}{A}$

Also $m_1$ & $m_2$ are slopes of two lines intersecting perpendicular to each other

$\therefore m_1{m}_2=-1=\frac{C}{A}$

$-1=\frac{y_1^2+b^2}{x_1^2-a^2}\Rightarrow a^2-x_1^2=y_1^2+b^2$

$y_1^2+x_1^2=a^2-b^2$

$\therefore$ locus of $P$ is $x^2+y^2=a^2-b^2$