Q75 / 155 · WBJEE 2019 · Mathematics

A variable line passes through the fixed point $(\alpha, \beta) .$ The locus of the foot of the perpendicular from the origin on the line is

  • Correct answer
    A
    $x^{2}+y^{2}-\alpha x-\beta y=0$
  • B
    $x^{2}-y^{2}+2 a x+2 \beta y=0$
  • C
    $\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0$
  • D
    $\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}}=1$

Solution

Let $(\alpha, \beta)$ be the given point, let $Q(x, y)$ be the foot of the perpendicular, and let $O$ be the origin. The line can have any direction $$ \angle P Q O=90^{\circ} $$ Point $Q$ lies on the circle having diameter $O P$. The locus of point $Q$. $$ (x-0)(x-\alpha)+(y-0)(y-\beta)=0 $$ $\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0$ $$ \Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0 $$

Hence, the correct answer is option: 'A'.

Relevant Questions