How you get y1/x1×(-q/p)=-1 in this question A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to a^2. Find the locus of the foot of perpendicular from the origin to this line

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How you get y1/x1×(-q/p)=-1 in this question A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to a^2. Find the locus of the foot of perpendicular from the origin to this line

How you get y1/x1×(-q/p)=-1 in this question A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to a^2. Find the locus of the foot of perpendicular from the origin to this line

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How you get y1/x1×(-q/p)=-1 in this question A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to a^2. Find the locus of the foot of perpendicular from the origin to this line

How you get y1/x1×(-q/p)=-1 in this question A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to a^2. Find the locus of the foot of perpendicular from the origin to this line

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Analytical Geometry

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship