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  • The ends A, B of a straight line segment of constant length ‘c’ slide upon the fixed rectangular axes OX and OY respectively. If the rectangle OAPB be completed, then the locus of the f

The ends A, B of a straight line segment of constant length ‘c’ slide upon the fixed rectangular axes OX and OY respectively. If the rectangle OAPB be completed, then the locus of the foot of the perpendicular, drawn from P to AB is \mathrm{x^{2 / 3}+y^{2 / 3}=k^{2 / 3}} where k =

Option: 1

\mathrm{C}


Option: 2

\mathrm{2C}


Option: 3

\mathrm{\frac{C}{2}}


Option: 4

\mathrm{\frac{3C}{2}}


Answers (1)

best_answer

Equation of AB is \mathrm{\frac{x}{a}+\frac{y}{b}=1}

It passes through (h, k), therefore,

\frac{\mathrm{h}}{\mathrm{a}}+\frac{\mathrm{k}}{\mathrm{b}}=1\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(1)

Equation of perpendicular to AB is

\begin{aligned} & \mathrm{b y-a x=b^2-a^2}\: \: \: \: \: \: \: \: \: \: \: \: \: .....(2) \\ \\& \mathrm{b x+a y=a b}\: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: .....(3) \end{aligned}

Solving , \mathrm{h=\frac{\mathrm{a}^3}{\mathrm{c}^2} \Rightarrow \mathrm{a}=\left(h c^2\right)^{1 / 3}}

\mathrm{k=\frac{b^3}{c^2} \Rightarrow b=\left(k c^2\right)^{1 / 3}}

Also \mathrm{a^2+b^2=c^2}

\therefore locus of \mathrm{(h, k)=x^{2 / 3}+y^{2 / 3}=c^{2 / 3}}

 

Posted by

Ritika Kankaria

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