Question

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

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

Answer 1

Answer: (A), (C)

$\frac{a}{\sqrt{2}}$


$xy=(a+x{)}^2$
$\Rightarrow y+x{y}^{\prime }=2(a+x)$
Now, $y^{\prime }=\pm 1$
$\Rightarrow y\pm x=2(a+x)$
$\Rightarrow \frac{(a+x{)}^2}{x}\pm x=2(a+x)$
$\Rightarrow \pm x=2(a+x)-\frac{(a+x{)}^2}{x}$
$\Rightarrow \pm x^2=(a+x)(x-a)$
$\Rightarrow \pm x^2=x^2-a^2$
$\Rightarrow 2x^2=a^2$
$\Rightarrow x=\pm \frac{a}{\sqrt{2}}$