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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 69 of 160

Marks: +1, -0

The slope of the normal to the curve

y=\frac{x}{x^{2}+1}

x=-4

is

$-\frac{289}{15}$
$-\frac{15}{16}$
$\frac{289}{15}$
$\frac{15}{16}$

Solution:

Slope of normal

=-\left( \frac{dx}{dy} \right)_{\{x_{1}, y_{1}\}}

Given curve,

y=\frac{x}{1+x^{2}}

, then

\frac{dy}{dx} =\frac{(1+x^{2})-x(2x)}{(1+x^{2})^{2}}=\frac{1-x^{2}}{(1+x^{2})^{2}}

\Rightarrow -\frac{dx}{dy} =-\frac{(1+x^{2})^{2}}{1-x^{2}}=\frac{(1+x^{2})^{2}}{x^{2}-1}

At

x=-4

\left( -\frac{dx}{dy} \right)_{x=-4} =\frac{((-4)^{2}+1)^{2}}{(-4)^{2}-1}

=\frac{(16+1)^{2}}{16-1}

=\frac{17^{2}}{15}=\frac{289}{15}

∴ Slope of normal to curve

y=\frac{x}{x^{2}+1}

, at

x=-4

is

\frac{289}{15}

.

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