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

Option c.)  ax²+by²+(a+b)z²=1.​

Given:

The locus of the point from which three mutually perpendicular lines can be drawn z=0

To Find:

When z= 0 and ax²+by²=1  to find where it intersects

Solution:

Let P(α,β,γ) be any point whose locus is to be found .Any line through

P(α,β,γ) is,

x-α/ l = y - β/m = z-γ/n

This meet the plane which lies on the given conic ,

α(α - lγ/n)^2 + b( β - (y-β/z-γ)γ)^2 =1

a(αz - xγ)^2 +b(βz - yγ)^2 = (z - γ)^2

Coeff of x^2 +Coeff of y^2+Coeff of z^2 =0

Therefore,

Locus of P(α,β,γ) is   ax²+by²+(a+b)z²=1.​

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