AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

It is given that 2y=3x−1 is tangent to y2=ax3+b at point
(1,1) with slope ‌

3

2

Differentiate the above equation with respect to x
2y(‌

dy

dx

)=3ax2
‌

dy

dx

=‌

3ax2

2y

At(1,1)
‌

dy

dx

=‌

3a(1)2

2(1)

=‌

3a

2

The slope is,
‌

dy

dx

=‌

3

2

‌

3a

2

=‌

3

2

a=1
The point (1,1) lies on y2=ax3+b
12=1×13+b
b=0
The point is (1,0)