Question

The abscissa of the point on the curve $yx=(x+a{)}^2$ at which the normal cuts off numerically equal intercepts on the coordinate axes is

• A. $\frac{2a}{\sqrt{2}}$
• B. $\frac{a}{\sqrt{2}}$
• C. $2\sqrt{a}$
• D. $a$

Answer 1

The correct option is C $2\sqrt{a}$

For normal to cut numerically equal intercepts there are few possibilities.
If it cuts intercepts of $(1,1)$, $(2,2)$ etc then slope $=-1$. If it cuts intercepts of $(-1,-1)$, $(-2,-2)$ etc. then slope = 1.

Therefore, for normal to cut numerically equal intercepts slope must be $=\pm 1$

$yx=(x+a{)}^2$
On differentiating the above equation w.r.t $x$, we have-
$y+x{y}^{{}^{\prime }}=2(x+a)$
$y^{{}^{\prime }}=\frac{2(x+a)-y}{x}=$ slope of tangent
Let $M_{\text{normal}}$ be the slope of normal $=\frac{-x}{2(x+a)-y}$
$=\frac{-x}{2(x+a)-\frac{(x+a{)}^2}{x}}[\because yx=(x+a{)}^2]$
$=\frac{-x^2}{2x(x+a)-(x+a{)}^2}=\pm 1$
For $M_{\text{normal}}=1$
$2x(x+a)-(x+a{)}^2=-x^2$
$\Rightarrow 2x^2-x^2-a^2=-x^2$
$\Rightarrow x=\frac{1}{\sqrt{2}}a$
For $M_{\text{normal}}=-1$
$-2x^2+x^2+a^2=-x^2$
No solution
$\therefore x=\frac{1}{\sqrt{2}}a$