TS EAMCET 5 May 2018 Shift 1 Solved Paper

The given equation is,
y2=ax3+b …… (1)
Differentiate the above equation with respect to x .
2y

dy

dx

=3ax2

dy

dx

=

3ax2

2y

Since, the slope is at (1, 2) then,

dy

dx

=

3a

4

…… (2)
The given tangent is y=2x
Slope of tangent =2 …… (3)
From equation (2) and equation (3),

3a

4

=2
a=

8

3

Substitute the value in equation (1),
y2=

8

3

x3+b
since, the curve passes through (1,2) then,
22=

8

3

(1)3+b
b=

4

3

Thus, (a,b)=(

8

3

,

4

3

)