Question

A ladder $12$ units long slides in a vertical plane with its ends in contact with a vertical wall and a horizontal floor along $x-$axis. Then the locus of a point on the ladder $4$ units from its foot, is

• A. $\frac{x^2}{64}+\frac{y^2}{32}=1$
• B. $\frac{x^2}{64}+\frac{y^2}{16}=1$
• C. $\frac{x^2}{32}+\frac{y^2}{32}=1$
• D. $\frac{x^2}{8}+\frac{y^2}{4}=1$

Answer 1

The correct option is B $\frac{x^2}{64}+\frac{y^2}{16}=1$

Given

Let $QR$ is a ladder, in which $Q\equiv (a,0)$ and $R\equiv (0,b)$

Given, $a^2+b^2={12}^2=144\dots \left(i\right)$
Clearly $P$ divides $QR$ internally in the ratio $1:2$
$\therefore$ By section formula, $P$ is of the form.
$P=(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
Here, $m=1,n=2$
$Q=(a,0)=(x_1,y_1)$ and $R=(0,b)=(x_2,y_2)$
$\therefore P=(x,y)=(\frac{1\times 0+2\times a}{1+2},\frac{1\times b+2\times 0}{1+2})$
i.e., $(x,y)=(\frac{2a}{3},\frac{b}{3})$
$\therefore x=\frac{2a}{3},y=\frac{b}{3}$
$\Rightarrow h=\frac{2a}{3}$ and $k=\frac{b}{3}$
$\Rightarrow a=\frac{3h}{2}$ and $b=3k$

From $\left(i\right),$ we get
$\frac{9h^2}{4}+9k^2=144$
$\Rightarrow \frac{x^2}{64}+\frac{y^2}{16}=1$