Question

The length of the normal to the curve $y=a\left(\frac{e^{-x/a}+e^{x/a}}{2}\right),$$a\in R-\left\{0\right\}$ at any point varies as the

• A. abscissa of the point
• B. ordinate of the point
• C. square of the abscissa of the point
• D. square of the ordinate of the point

Answer 1

The correct option is D square of the ordinate of the point

$y=a\left(\frac{e^{-x/a}+e^{x/a}}{2}\right)$
$\Rightarrow \frac{dy}{dx}=\left(\frac{e^{x/a}-e^{-x/a}}{2}\right)$
Length of the normal at any point
$=y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$
$=y\sqrt{1+{\left(\frac{e^{x/a}-e^{-x/a}}{2}\right)}^2}$
$=y\left(\frac{e^{x/a}+e^{-x/a}}{2}\right)$
$=y\left(\frac{y}{a}\right)=\frac{y^2}{a}$
Hence, the length of the normal at any point varies as the square of the ordinate of the point.