Question

Find the equation of the normal to the curve y=x² at (1,1)​

Answer 1

Given :

• Equation of curve : y=x²

To find :

• Equation of the normal to the curve y=x² at (1,1)

Concept used :

$\sf m_1 \times m_2 = - 1$

Where :-

• m1 = slope of tangent
• m2 = slope of normal

Solution :

To find Slope of tangent we have to differentiate the given equation of curve. (dy/dx at (1,1) will be the required slope of tangent )

$\implies \sf y = {x}^{2}$

Differentiating both sides :-

$\implies \sf \dfrac{dy}{dx} = \dfrac{d( {x}^{2}) }{dx}$

We know that :-

$\boxed { \bf\dfrac{d( {x}^{n}) }{dx} = n \times {x}^{n - 1} }$

$\implies \sf \dfrac{dy}{dx} = 2x$

Now put x = 1

$\implies \sf \dfrac{dy}{dx} = 2 \times 1$

$\implies \sf \dfrac{dy}{dx} = 2$

Therefore , slope of tangent of given curve at (1,1) = 2

Now we have to find out slope of normal to the given curve.

We know that :-

$\sf m_1 \times m_2 = - 1$

Put m1 = 2

$\implies \sf 2 \times m_2 = - 1$

$\implies \sf m_2 = \dfrac{ - 1}{2}$

Now , normal to the curve y= x² straight line passing through point (1,1) with slope -1/2.

Standard equation of straight line passing through point (x1,y1) and has slope m :-

$\bf y - y_1 = (x - x_1)m$

Therefore, equation of normal :-

$\implies \sf y -1 = (x -1) \bigg(\dfrac{ - 1}{2} \bigg )$

$\implies \sf 2(y -1 )= (x -1) ({ - 1})$

$\implies \sf 2y -2= 1 - x$

$\implies \sf 2y -2 + x - 1=0$

$\implies \sf 2y + x - 3=0$

Answer :

• 2y+x-3 =0