The abscissae of the point on the curve x y=(a+x)^2, the normal at wh…

We have,

$x y=(a+x)^2$           ……..(i)

$\Rightarrow y=x+2 a+\frac{a^2}{x} \Rightarrow \frac{d y}{d x}=1-\frac{a^2}{x^2}$

Let $P\left(x_1, y_1\right)$ be a point on the curve (i), where the normal cuts off numerically equal intercepts from the coordinate axes.

Then,

$\frac{-1}{\left(\frac{d y}{d x}\right)_P}= \pm 1 \Rightarrow\left(\frac{d y}{d x}\right)_P= \pm 1 \Rightarrow 1-\frac{a^2}{x_1{ }^2}= \pm 1 $

$\Rightarrow 1-\frac{a^2}{x_1^2}=1 \text { or, } 1-\frac{a^2}{x_1{ }^2}=-1 $

$\Rightarrow \frac{a^2}{x_1{ }^2}=0 \text { or, } x_1= \pm \frac{a}{\sqrt{2}} \Rightarrow x_1= \pm \frac{a}{\sqrt{2}}$