Question

Determine a and b such that slope of curve 2y^3 = ax^2 + b at (1,-1) is same as the slope of x+y =0



3,1

2,1

-3,1

1,2​

Answer 1

Answer:

The values of a and b are $a=-3\:and\:b=1$

Step-by-step explanation:

The given curve is $2y^{3}=ax^{2}+b$

It is said that at the point (1,-1) the slope of the curve is equal to the slope of line $x+y=0$

To find the slope,we shall differentiate the curve as shown below

$2y^{3}=ax^{2}+b\\= > 6y^{2}\frac{dy}{dx} =2ax\\= > m=\frac{ax}{3y^{2}}\\m_{(1,1)}=\frac{a}{3}$

The slope of the line $x+y=0$ is $-1$

equating them together we get $\frac{a}{3} =-1= > a=-3$

Hence the curve becomes $2y^{3}=-3x^{2}+b$

Substituting the point we get $-2=-3+b= > b=1$

Therefore,

The values of a and b are $a=-3\:and\:b=1$

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