Question

Let f : $R\to R$ be a continuous function satisfying $(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)$ for all $x,yϵR$ and $f\left(1\right)=1.$

Locus of the mid-point of a segment of the tangent intercepted between the coordinate axes, drawn at any point on the curve $\left(f\right(x){)}^{2/9}+(f\left(y\right){)}^{2/9}=1$ is

• A. A straight line
• B. A parabola
• C. A circle
• D. An ellipse

Answer 1

Answer: (C)

A circle

$(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2)$

$\Rightarrow (x-y)f(x+y)-(x+y)f(x-y)=4xy(x-y)(x+y)$

$\Rightarrow \frac{f(x+y)}{x+y}-\frac{f(x-y)}{x-y}=4xy$

Substitute $x+y=t,x-y=u\Rightarrow t^2-u^2=4xy$

The equation becomes:

$\Rightarrow \frac{f\left(t\right)}{t}-\frac{f\left(u\right)}{u}=t^2-u^2$

Using $u=1$ in the above equation, we get

$\Rightarrow \frac{f\left(t\right)}{t}-\frac{f\left(1\right)}{1}=t^2-1$

Given that $f\left(1\right)=1$

$\Rightarrow \frac{f\left(t\right)}{t}-\frac{1}{1}=t^2-1$

$\Rightarrow f\left(t\right)=t^3$

$\Rightarrow f\left(x\right)=x^3$.....(1)

So, the equation of the curve $\left(f\right(x){)}^{2/9}+(f\left(y\right){)}^{2/9}=1$ can be written as

$x^{2/3}+y^{2/3}=1$.....(1)

Differentiating with respect to 'x',

$\frac{2}{3}{x}^{-1/3}+\frac{2}{3}{y}^{-1/3}\frac{dy}{dx}=0\frac{dy}{dx}=-x^{-1/3}{y}^{-1/3}$
At any point $(x_1,y_1)$ on the curve, the equation of the tangent is

$y-y_1=-{x_1}^{-1/3}{y_1}^{-1/3}(x-x_1)$

$\frac{x}{{x_1}^{1/3}}+\frac{y}{{y_1}^{1/3}}={x_1}^{2/3}+{y_1}^{2/3}\Rightarrow \frac{x}{{x_1}^{1/3}}+\frac{y}{{y_1}^{1/3}}=1$

From the above equation, the X-intercept is ${x_1}^{1/3}$ and the Y-intercept is ${y_1}^{1/3}$

Hence, coordinates of the mid-point $(h,k)$ of the tangent line intercepted between coordinate axes are

$h=\frac{{x_1}^{1/3}}{2},k=\frac{{y_1}^{1/3}}{2}$

Squaring and adding we get

$h^2+k^2=\frac{{x_1}^{2/3}+{y_1}^{2/3}}{4}=\frac{1}{4}$

Hence, locus of $(h,k)$ is a circle with centre at origin and radius $\frac{1}{2}$