Calculus Problem/ Derivatives?

Question

For what values of a and b is the line -2x+y=b tangent to the curve y=ax^2 when x=–4?

Answer 1

The curve equation is y=ax² .

Given tangent line is -2x+y=b  ---------->(1)

To find out tangent for the above curve , equate the first derivative of  curve equation to zero .

y' = 2ax              [ since the derivative of x^n = n x^n-1 ]

y-component   y = a(-4)² = 16 a                    [given x = -4 ]

So the point is (-4 , 16a )

Now we have to find out the tangent at x = -4 .

y' = 2a(-4)

y' = -8a 

So the slope of tangent is -8a and the point is (-4 , 16a ) .

So the equation of tangent is with point - slope form is 

y - y1 = m(x - x1)

y - 16a = -8a ( x - (-4))

y - 16a = -8ax  -32a 

8ax + y +16a = 0  ---------->(2)

Compare (1) and (2)

-2 = 8a   and -b = 16a 

a = -1/4    and -b = 16(-1/4)

a = -1/4    and -b = -4

So the the value of a = -1/4 .

So the the value of b = -4 .