Question

slope ds normal to the curve 2y=3-x2at (1,1)is
​

Answer 1

The curve is

2y=

3−

x 2

Slope

=

2y

′

=

−2x

or

d x

d y

=

−x

Slope at (1,1) is

d x

d y

=

−1

Equation of normal is

y−

1=

− 1

− 1

(x−1)

⇒ x− 1− y+ 1= 0

⇒ x− y= 0 is the equation of the normal.

Answer 2

Answer:

1

Step-by-step explanation:

2y=3-$x^{2}$

Differentiating the equation, we get

$2\frac{dy}{dx} =-2x\\\frac{dy}{dx} = -x$(Slope of the tangent)

Slope of normal=$\frac{-1}{Slope of tangent}$

$\frac{1}{x} \\=\frac{1}{1}\\\\=1$