Question

A rod $AB$ of length $15cm$ rests in between two coordinates axes in such a way that the end point A lies on x-axis and end point $B$ lies on y-axis. A point $P(x,y)$ is taken on the rod in such a way that $AP=6cm.$ show that the locus of $p$ is an ellipse.

Answer 1

Let $AB$ be rod & $P(x,y)$ be any point on rod such that $AP=6cm,AB=15cm$ & $BP=9cm$

Let $A(a,0)$ & $B(0,b)$ be the co-ordinates of point $A$ and $B$

$P(x,y)$ divides $AB$ in ratio $6:9$

$\therefore$ By section formula

$x=\frac{2\left(0\right)+3\left(a\right)}{2+3}$ & $y=\frac{\left(2b\right)+3\left(0\right)}{2+3}$

$x=\frac{3a}{5}$ & $y=\frac{2b}{5}$

$a=\frac{5x}{3}$ & $b=\frac{5y}{2}$......(1)

In $△OAB$,

$(OA{)}^2+(OB{)}^2=(AB{)}^2$

$a^2+b^2=(15{)}^2$

${\left(\frac{5x}{3}\right)}^2+{\left(\frac{5y}{2}\right)}^2=225$

$\frac{25x^2}{9}+\frac{25y^2}{4}=225$

$\frac{x^2}{9}+\frac{y^2}{4}=9$

$\frac{x^2}{81}+\frac{y^2}{36}=1$

$\frac{x^2}{(9{)}^2}+\frac{y^2}{(6{)}^2}=1$ which represents an ellipse.