Question

Let tangent at a point P on the curve $x^{2m}=y^{\frac{n}{2}}=a^{\frac{4m+n}{2}}$ meets the x-axis and y-axis at A and B respectively, If AP:PB is $\frac{n}{\lambda m}$, where P lies between A and B, then find the value of $\lambda$

• A. $4$
• B. $3$
• C. $-4$
• D. $-3$

Answer 1

The correct option is A $4$

$x^{2m}{y}^{\frac{n}{2}}=a^{\frac{4m+n}{2}}$
$2m\ln x+\frac{n}{2}\ln y=\frac{4m+n}{2}\ln a$
$\frac{\left(2m\right)}{x}+\left(\frac{n}{2}\right)\frac{1}{y}\frac{dy}{dx}=0$
$\frac{dy}{dx}=\left(\frac{-2m}{x}\right)\frac{2y}{n}$
Let $P$ be the point $(x_1,y_1)$
So equation of tangent is
$y-y_1=(\frac{-4m}{n}.\frac{y_1}{x_1})(x-x_1)$.......(i)
Let Tangents meets the axes at $A$ and $B$ respectively
$A=(\frac{4m+n}{4m}{x}_1,0)B=(0,\frac{4m+n}{n}.y_1)$
Let $P$ divides $AB$ in ratio $\mu :1$
$\therefore x_1=\frac{\mu .0+1.\left(\frac{4m+n}{4m}\right)x_1}{\mu +1}$

$\therefore 4m+n=4m(\mu +1)$
Also $\mu (4m+n)=n(\mu +1)$

$\therefore \mu =\frac{n}{4m}$
$\therefore \lambda =4$