Practicing Success
The abscissae of the point on the curve $x y=(a+x)^2$, the normal at which cuts off numerically equal intercepts from the axes of coordinates, is (a) $\frac{a}{\sqrt{2}}$ |
(a), (c) (a), (d) (b), (d) (b), (c) |
(a), (d) |
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 $ $\Rightarrow \frac{a^2}{x_1{ }^2}=0 \text { or, } x_1= \pm \frac{a}{\sqrt{2}} \Rightarrow x_1= \pm \frac{a}{\sqrt{2}}$ |