The equations of the normals to the curve f x - x -1- x 2 at the points where the tangents make the angle of -4 with the positive direction of X - axis are-A- x-y-0 B- x-y-3-2C- x-y-2 -2D - x - y -3-2

The correct options are A x + y = 0 B x + y = nbsp; √(3)/2 D x + y = nbsp; √(3)/2dy/dx = 1+x^2/(1 x^2)^2 = 1 ⇒ x^4 3x^2 = 0 ⇒ x = 0, ±√(3) nbsp; n ...

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Byju's Answer

Standard XII

Mathematics

Tangent of a Curve y =f(x)

The equations...

Question

The equations of the normals to the curve $f (x) = \frac{x}{1 - x^{2}}$ at the points where the tangents make the angle of $\frac{π}{4}$ with the positive direction of $x$ - axis are:

x + y = 0

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x + y = \frac{\sqrt{3}}{2}

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x + y = 2 \sqrt{2}

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x + y = - \frac{\sqrt{3}}{2}

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Solution

The correct options are
A $x + y = 0$
B $x + y = \frac{\sqrt{3}}{2}$
D $x + y = - \frac{\sqrt{3}}{2}$
$\frac{d y}{d x} = \frac{1 + x^{2}}{(1 - x^{2})^{2}} = 1 \Rightarrow x^{4} - 3 x^{2} = 0 \Rightarrow x = 0, \pm \sqrt{3}$

$x = 0, y = 0$ ;
$x = \sqrt{3}, y = - \frac{\sqrt{3}}{2}$ ;
$x = - \sqrt{3}, y = \frac{\sqrt{3}}{2}$
So the equation of the normals: $(y - 0) = - 1 (x - 0)$
$(y + \frac{\sqrt{3}}{2}) = - 1 (x - \sqrt{3})$
$(y - \frac{\sqrt{3}}{2}) = - 1 (x + \sqrt{3})$

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The equations of the normals to the curve f(x)=x1−x2 at the points where the tangents make the angle of π4 with the positive direction of x - axis are:

The correct options are A x+y=0 B x+y=√32 D x+y=−√32dydx=1+x2(1−x2)2=1⇒x4−3x2=0⇒x=0,±√3 x=0,y=0; x=√3, y=−√32; x=−√3, y=√32 So the equation of the normals: (y−0)=−1(x−0) (y+√32)=−1(x−√3) (y−√32)=−1(x+√3)

The equations of the normals to the curve $f (x) = \frac{x}{1 - x^{2}}$ at the points where the tangents make the angle of $\frac{π}{4}$ with the positive direction of $x$ - axis are: