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

 

Option: 1

\mathrm{c}


Option: 2

\mathrm{2c}


Option: 3

\mathrm{\frac{c}{2}}


Option: 4

\mathrm{8c^{3}}


Answers (1)

best_answer

Let \mathrm{A \equiv(a, 0) B \equiv(0, b) \text\ { then }\ P \equiv(a, b)}
\mathrm{\begin{aligned} & \text { Since } A B=c \quad \sqrt{a^2+b^2}=c \\ & \text { or } \quad a^2+b^2=c^2 \ \ \ \ \ ...........\left ( 1 \right )\\ & \text { and let } Q \equiv\left(x_1, y_1\right) \\ & \triangle \quad P Q \perp A B \\ & \therefore \text { Slope of } P Q \times \text { slope of } A B=-1 \end{aligned}}

\mathrm{\Rightarrow\left(\frac{\mathrm{b}-\mathrm{y}_1}{\mathrm{a}-\mathrm{x}_1}\right) \times\left(\frac{\mathrm{b}-0}{0-\mathrm{a}}\right)=-1}
\mathrm{\begin{aligned} & \Rightarrow \mathrm{ax}_1-\mathrm{by}_1=\mathrm{a}^2-\mathrm{b}^2 \ldots \ldots \text { (2) } \\ & \Delta \text { Equation of } \mathrm{AB} \text { is } \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1 \end{aligned}}
\mathrm{but\ Q\ lies\ on\ A B\ then\ \frac{x_1}{a}+\frac{y_1}{b}=1}
\mathrm{\Rightarrow \quad \mathrm{bx}_1+\mathrm{ay}_1=\mathrm{ab}\ \ \ \ \ .........\left ( 3 \right )}
from \mathrm{\left ( 2 \right )} and \mathrm{\left ( 3 \right )} we get \mathrm{\mathrm{x}_1=\frac{\mathrm{a}^3}{\mathrm{a}^2+\mathrm{b}^2}, \mathrm{y}_1=\frac{\mathrm{b}^3}{\mathrm{a}^2+\mathrm{b}^2}}
Now \mathrm{\mathrm{x}_1^{2 / 3}+\mathrm{y}_1^{2 / 3}=\frac{\left(\mathrm{a}^2+\mathrm{b}^2\right)}{\left(\mathrm{a}^2+\mathrm{b}^2\right)^{2 / 3}}}
\mathrm{=\left(a^2+b^2\right)^{1 / 3}}
\mathrm{=c^{2/3} \ \ \ \ \\ \ \ \left [ from\ \left ( 1 \right ) \right ]}
Hence required locus is \mathrm{x^{2 / 3}+y^{2 / 3}=c^{2 / 3}}

Posted by

Rakesh

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