Question

Find the value of a and b for the curve x^2y+ay^2=b if the point (1,1) is on its graph and the tangent line at (1,1) has the equation 4x+3y=7

Answer 1

By plugging (1, 1) into the equation, we get 1+a=b.
Do you know how to find the tangent line, or even how to find the derivative of the equation x^2y+ay^2=b?

Answer 2

Hint: you can set the derivative of your function at \((1,1)\) equal to the slope of the tangent line.

Answer 3

This can be solved because of the included bit about the tangent line, because now you have enough information to create 2 simultaneous equations.

You have, x^(2y) + ay^2 = b, and the other is found by finding its derivative.

2y*x^(2y - 1) + 2ay * dy/dx = 0

And we know what x, y, and dy/dx are.
x is 1 and y is 1 from the coordinate (1,1)
dy/dx can be found with the tangent equation, by expressing it in slope-intercept form.

4x + 3y = 7
3y = 7 - 4x
y = 7/3 - (4/3)x

So, dy/dx is 4/3.

Now subsitute x, y and dy/dx into each equation, and simplify it.

x^(2y) + ay^2 = b
(1)^(2(1)) + a(1)^2 = b
1 + a = b

2y*x^(2y - 1) + 2ay * dy/dx = 0
2(1)*(1)^(2(1) - 1) + 2a(1)(4/3) = 0
2 + 8a/3 = 0

You can now solve for a.

2 + 8a/3 = 0
8a/3 = -2
8a = -6
a = -3/4

And now back substituting for a to find b.

1 + a = b
1 + (-3/4) = b
b = 1/4

Therefore,

a = -3/4
b = 1/4