Question

Tangent at any point on the curve $x^2{y}^3=a^5$ has an intercept between the axes. This intercept is divided in the ratio $m:n,$ such that $m>n$ and $m,n$ are co-prime, then $4m+2n=$

Answer 1

Let $P(x_1,y_1)$ be a point on the curve $x^2{y}^3=a^5$
$\Rightarrow x^2\times 3y^2{y}^{\prime }+2x{y}^3=0$
$\Rightarrow \frac{dy}{dx}=-\frac{2y}{3x}$

Equation of tangent at $(x_1,y_1)$ is
$y-y_1=-\frac{2y_1}{3x_1}(x-x_1)$
$\Rightarrow \frac{3(y-y_1)}{y_1}=-\frac{2(x-x_1)}{x_1}$
$\Rightarrow \frac{2x}{x_1}+\frac{3y}{y_1}=5$

Let $A$ and $B$ be the point of intersection on $x$ and $y$ axes.
Then, coordinates are $A(\frac{5x_1}{2},0)$ $B(0,\frac{5y_1}{3})$
Now, $PA=\sqrt{{(\frac{5x_1}{2}-x_1)}^2+{(0-y_1)}^2}$
$\Rightarrow PA=\sqrt{\frac{9x_1^2}{4}+y_1^2}=\sqrt{\frac{9x_1^2+4y_1^2}{4}}$

$PB=\sqrt{{(0-x_1)}^2+{(y_1-\frac{5y_1}{3})}^2}$
$\Rightarrow PB=\sqrt{x_1^2+\frac{4y_1^2}{9}}=\sqrt{\frac{9x_1^2+4y_1^2}{9}}$
So the ratio will be
$\frac{PA}{PB}=\sqrt{\frac{9}{4}}\Rightarrow \frac{3}{2}=\frac{m}{n}$

Hence, the value of $4m+2n=16$