Question

Two ends A & B of a straight line segment of constant length 'C' slide upon the fixed rectangular axes OX & OY respectively. If the rectangle OAPB is completed. Then find locus of the foot of the perpendicular drawn from P to AB.

• A. $x^{\frac{2}{3}}+y^{\frac{2}{3}}=c^{\frac{2}{3}}$
• B. $x^{\frac{2}{3}}+y^{\frac{2}{3}}=c^{\frac{1}{3}}$
• C. $x^{\frac{1}{3}}+y^{\frac{1}{3}}=c^{\frac{2}{3}}$
• D. $x^{\frac{1}{3}}+y^{\frac{1}{3}}=c^{\frac{1}{3}}$

Answer 1

The correct option is A $x^{\frac{2}{3}}+y^{\frac{2}{3}}=c^{\frac{2}{3}}$

Let the coordinates of A and B be (a,0) Ans (0,b) respectively,As the rod slides the values of a and b change , so a and b are variables,

Also the point P is formed by completing the rectangle OAPB so the co ordinates of P are (a,b)

let Q(h,k) be the foot of perpendicular of perpendicular from P on AB.As the length of the rod is constant c, then

$a^2+b^2=c^2..................\left(i\right)$

Slope of the line AB=$-\frac{b}{a}$

Slope of the line AB=$-\frac{k}{a-h}$

Slope of the line AB=$-\frac{b-k}{h}$

As A,Q,B lie on the same line,therfore

$-\frac{b}{a}=-\frac{k}{a-h}=\frac{h-k}{h}$

$⟹a-h=\frac{a}{b}k$ and $h-k=\frac{b}{a}h$

Also slope of PQ=$\frac{b-k}{a-h}$

Now ,PQ is perpendicular to AB ,then,

$\frac{b-k}{a=h}\times (-\frac{b}{a})=-1$

$⟹\frac{\frac{b}{a}k}{\frac{a}{b}h}\times \frac{b}{a}=1$

$⟹a^{3}k=b^{3}h....................\left(ii\right)$

Also using the intercept form ,the equation of line AB is

$\frac{x}{a}+\frac{y}{b}=1$

As Q(h,k) lies on AB,therfore

$\frac{h}{a}+\frac{k}{b}=\mathrm{1........................}\left(iii\right)$

Putting $k=\frac{b^3}{a^3}h$ from (ii) in equation (iii),we get

$h\frac{a^2+b^2}{a^3}=1$

$⟹h\left(\frac{c^2}{a^3}\right)=1⟹h=\frac{a^2}{c^2}⟹a=(h{c}^2{)}^{\frac{1}{3}}$

So, $k=\frac{b^3}{a^3}h=\frac{b^3}{a^3}\left(\frac{a^3}{c^2}\right)=\frac{b^3}{c^2}⟹b=(k{c}^2{)}^{\frac{1}{3}}$

Putting values of b and a in (i),we get

$(h{c}^2{)}^{\frac{1}{3}}+(k{c}^2{)}^{\frac{1}{3}}=c^2$

$(h{)}^{\frac{2}{3}}+(k{)}^{\frac{2}{3}}=c^{\frac{2}{3}}$

Therfore reuired locus of Q is,

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=c^{\frac{2}{3}}$