Determine the slope of the line tangent to the graph of the function 3(x2+y2)2=100(x2−y2) at the point (4,2) . (1 point) Responses –411 – 4 11 –211 – 2 11 2548 25 48 8885 88 85 17685
To find the slope of the tangent line, we will need to find the derivative of the given equation, set up an equation using the given point (4,2), and solve for the slope.
Taking the derivative of the given equation, we get:
d/dx[3(x^2+y^2)^2]=d/dx[100(x^2-y^2)]
Using the chain rule, the derivative of (x^2+y^2)^2 is:
2(x^2+y^2)(2x+2yy')
The derivative of 100(x^2-y^2) is:
200x-200yy'
Now, we can set up an equation using the given point (4,2):
2(4^2+2^2)(2(4)+2(2)y') = 200(4)-200(2)y'
2(16+4)(8+4y') = 800-400y'
2(20)(8+4y') = 800-400y'
40(8+4y') = 800-400y'
Expanding,
320 + 160y' = 800 - 400y'
Rearranging,
160y' + 400y' = 800 - 320
560y' = 480
y' = 480/560
y' = 6/7
Therefore, the slope of the tangent line at the point (4,2) is 6/7.