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  • A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to <img alt="\mathrm{a^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7

A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to \mathrm{a^2}. The locus of the foot of the perpendicular from the origin to this line is \mathrm{\left(x^2+y^2\right)^2\left(x^{-2}+y^{-2}\right)=k}. Find k

Option: 1

\mathrm{k=a}


Option: 2

\mathrm{k=-a^{2}}


Option: 3

\mathrm{k=a^{2}}


Option: 4

\mathrm{k=2a^{2}}


Answers (1)

best_answer

Let P (h, k) be the foot of the perpendicular of the line,

so the slope of variable line must be \mathrm{-\frac{h}{k}}

Its equation should be \mathrm{y-k=-\frac{h}{k}(x-h)}

i.e. \mathrm{\frac{x}{\frac{(h^{2}+k^{2})}{h}}+\frac{y}{\frac{(h^{2}+k^{2})}{k}}}=1

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

Posted by

Rishabh

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