Question

The tangent at a point $P$ on the curve $x^m{y}^n=a^{m+n}$ meets the $x$-axis at $A$ and $y$-axis at $B$. The triangle formed by $AOB$ has centroid at $Q$, If $O$ be the origin, then

• A. $AP:PB=n:m$
• B. $AP:PB=m:n$
• C. locus of $Q$ is $\left(3xm{)}^m\right(3yn{)}^n=\left[a\right(m+n){]}^{m+n}$
• D. locus of $Q$ is $\left(3xm{)}^n\right(3yn{)}^m=\left[a\right(m+n){]}^{m+n}$

Answer 1

Answer: (A), (C)

The correct options are
A $AP:PB=n:m$
C locus of $Q$ is $\left(3xm{)}^m\right(3yn{)}^n=\left[a\right(m+n){]}^{m+n}$

$x^m{y}^n=a^{m+n}\Rightarrow m\ln x+n\ln y=(m+n)\ln a$
Differentiating w.r.t. $x$
$\frac{m}{x}+\frac{n}{y}{y}^{\prime }=0\Rightarrow y^{\prime }=-\frac{my}{nx}$
So, equation of the tangent at $P(x,y)$ is
$Y-y=-\frac{my}{nx}(X-x)$
Tangent meets the $x$-axis at $A$ and $y$-axis at $B$.
$\therefore A=(\frac{(m+n)x}{m},0)B=(0,\frac{(m+n)y}{n})$

Let $P$ divides $AB$ internally in the ratio of $r:1$
Then $x=\frac{r\cdot 0+1\cdot \frac{(m+n)x}{m}}{r+1}$
$\Rightarrow (r+1)x=(1+\frac{n}{m})x\Rightarrow r=\frac{n}{m}\therefore AP:PB=n:m$

Let coordinates of $Q$ is $(h,k)$
$\therefore h=\frac{(m+n)x}{3m}$
and $k=\frac{(m+n)y}{3n}$
$\Rightarrow (3hm{)}^m\cdot (3kn{)}^n=(m+n{)}^{m+n}{x}^m{y}^n\Rightarrow (3xm{)}^m(3yn{)}^n=[a(m+n){]}^{m+n}$