WBJEE · Maths · Straight Lines
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
- 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 \(\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}}=1\)
Answer & Solution
Correct Answer
(A) \(x^{2}+y^{2}-\alpha x-\beta y=0\)
Step-by-step Solution
Detailed explanation
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 \(Q\).…
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