Question

3. Equation of normal to the curve
3ay^2=x^2(x+a) at (2a, 2a) is....
O A. 3x+y=4
B. x+2y=5
O C. 4x+3y=14
O D. 3x+4y=14a​

Answer 1

Answer:

The curve is 2y=3−x

2

Slope=2y

′

=−2x or

dx

dy

=−x

Slope at (1,1) is

dx

dy

=−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.