Question:
The equation of the normal to the curve $y(1 + x^2) = 2 - x$ where the tangent crosses $x-axis$ is
The equation of the normal to the curve $y(1 + x^2) = 2 - x$ where the tangent crosses $x-axis$ is
Updated On: May 14, 2024
- $5 x - y - 10 = 0$
- $ x - 5y - 10 = 0$
- $5 x + y + 10 = 0$
- $ x + 5 y + 10 = 0$
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The Correct Option is A
Solution and Explanation
We have, $y\left(1+x^{2}\right)=2-x\dots$(i)
Put $y=0 \Rightarrow x=2 [\because$ tangent crosses $X$ -axis]
On differentiating E (i) w.r.t. $x$, we get
$\frac{d y}{d x}\left(1+x^{2}\right)+2 x y=-1$
$ \Rightarrow \frac{d y}{d x}=\frac{-1-2 x y}{1+x^{2}}$
$\therefore\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{-1-0}{1+4}=\frac{-1}{5}$
So, the slope of normal is $5$ .
$\therefore$ Equation of the normal at $(2,0)$ is
$ y-0 =5(x-2) $
$\Rightarrow y =5 x-10 $
$\Rightarrow 5 x-y-10 =0$
Put $y=0 \Rightarrow x=2 [\because$ tangent crosses $X$ -axis]
On differentiating E (i) w.r.t. $x$, we get
$\frac{d y}{d x}\left(1+x^{2}\right)+2 x y=-1$
$ \Rightarrow \frac{d y}{d x}=\frac{-1-2 x y}{1+x^{2}}$
$\therefore\left(\frac{d y}{d x}\right)_{(2,0)}=\frac{-1-0}{1+4}=\frac{-1}{5}$
So, the slope of normal is $5$ .
$\therefore$ Equation of the normal at $(2,0)$ is
$ y-0 =5(x-2) $
$\Rightarrow y =5 x-10 $
$\Rightarrow 5 x-y-10 =0$
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