Question

The abscissa of a point on the curve $xy=(a+x{)}^2$, 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)

The correct options are
A $-\frac{a}{\sqrt{2}}$
C $\frac{a}{\sqrt{2}}$

Given,

$xy=(a+x{)}^2$

differentiating the above equation, we get,

$\Rightarrow y+x{y}^{\prime }=2(a+x)$

substituting and solving the above equation, we get,

$\therefore y^{\prime }=\pm 1$

$y\pm x=2(a+x)$

$\frac{(a+x{)}^2}{x}\pm x=2(a+x)$

$\Rightarrow \pm x=2(a+x)-\frac{(a+x{)}^2}{x}$

$\pm x^2=(2+x)[x-a]$

$\pm x^2=x62-a^2$

$\Rightarrow 2x^2=a^2$

$\therefore x=\pm \frac{a}{\sqrt{2}}$