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  • The two points A and B move on the positive direction of x-axis and y-axis respectively, such that  <img alt="\mathrm{OA + OB = K}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

The two points A and B move on the positive direction of x-axis and y-axis respectively, such that  \mathrm{OA + OB = K}. Find ‘a’ if the locus of the foot of the perpendicular from the origin O on the line AB is \mathrm{ (x + y)(x^{2} + y^{2}) = 2axy.}

Option: 1

\mathrm{K}


Option: 2

\mathrm{2K}


Option: 3

\mathrm{\frac{3K}{2}}


Option: 4

\mathrm{\frac{K}{2}}


Answers (1)

best_answer

Let the equation of AB be \mathrm{\frac{x}{a}+\frac{y}{b}=1}\: \: \: \: \: \: \: \: \: \: ...(1)

Given, \mathrm{a+b=k}\: \: \: \: \: \: \: \: ...(2)

New \mathrm{m_{A B} \quad m_{O M}=-1}

\mathrm{ah=bk}\: \: \: \: \: \: \: \: \: \: \: \: \: ....(3)

from (2) and (3),

\mathrm{a=\frac{k K}{h+k} \text { and } b=\frac{h K}{h+k}}

Putting in (1),

\mathrm{\frac{x(h+k)}{k K}+\frac{y(h+k)}{h K}=1}

\therefore locus of (h,k) is \mathrm{(x+y)\left(x^2+y^2\right)=K x y}

Posted by

Ajit Kumar Dubey

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