Question

Prove that the locus of the point from which three mutually perpendicular lines can be drawn to intersect the conic z=0,ax²+by²=1 is given by ax²+by²+(a+b)z²=1.

Answer 1

The locus of the point from which three mutually perpendicular lines can be drawn to intersect the conic is given by the equation ax² + by² + (a + b)z² = 1.

Step-by-step explanation:

To prove that the locus of the point from which three mutually perpendicular lines can be drawn to intersect the conic, we can start by considering the equations of the three lines passing through the point (x, y, z):

1. Line 1: z = 0
2. Line 2: ax² + by² = 1
3. Line 3: (a + b)z² = 1

By solving these equations simultaneously, we can show that the locus of the point is given by the equation ax² + by² + (a + b)z² = 1.