Question

In the curve $x^a{y}^b=k^{a+b},(ab>0)$ then the portion of the tangent intercepted between coordinate axes is divided at its point of contact into segments which are in constant ratio.

• A. True
• B. False

Answer 1

The correct option is A True

Let P$(x_1,y_1)$ be the point of contact of the tangent.

Given $x^a{y}^b=k^{(}a+b)$

$⟹alogx+blogy=(a+b)logk$

$⟹\frac{a}{x}+\frac{b}{y}.\left(\frac{dy}{dx}\right)=0$

$⟹\frac{dy}{dx}=–\frac{ay}{bx}$

Equation of the tangent at P is

$y-y_1=(-\frac{a{y}_1}{b{x}_1})(x-x_1)-\left(1\right)$

Put y=0 in (1) , we get, x=($\frac{a+b}{a})x_1$

$\to A=\left(\right(\frac{a+b}{a})x_1,0)$

put x=0in (1) we get , $y=\left(\frac{a+b}{a}\right)y_1$

thus $B=(0,(\frac{a+b}{b}\left)y_1\right)$

Let P divide AB in the ratio $\lambda :1$

$⟹P=(\frac{\left(\frac{a+b}{a}\right)x_1}{\lambda +1},\frac{\lambda \left(\frac{a+b}{b}\right)y_1}{\lambda +1})$

Thus,

$x_1=\frac{\left(\frac{a+b}{a}\right)x_1}{\lambda +1},y_1=\frac{\lambda \left(\frac{a+b}{b}\right)y_1}{\lambda +1}$

$⟹\lambda +1=\left(\frac{a+b}{a}\right),\lambda +1=\left(\frac{a+b}{b}\right)$

$\lambda =\frac{b}{a}or\frac{a}{b}$

Therefore P divides AB in the ratio a : b.