Final answer:
The equation of the line tangent to the curve 2(x²+y²)²=25(x²-y²) at the point (-3,-1) is y = (-9/5)x - 32/5.
Step-by-step explanation:
To find the equation of the line tangent to the curve 2(x²+y²)²=25(x²-y²) at the point (-3,-1), we need to find the slope of the tangent line at that point.
First, we differentiate both sides of the equation with respect to x.
The derivative of 2(x²+y²)² is 4(x²+y²)(2x+2y⋅dy/dx), and the derivative of 25(x²-y²) is 50x-50y⋅dy/dx.
Setting the derivative equal to the slope of the tangent line, we have 4(x²+y²)(2x+2y⋅dy/dx) = 50x-50y⋅dy/dx.
Substituting the coordinates (-3,-1) into this equation, we can solve for dy/dx, which turns out to be -9/5.
Therefore, the equation of the tangent line is y = (-9/5)x + b.
To find the value of b, substitute the point (-3,-1) into this equation and solve for b.
Plugging in the values, we get -1 = (-9/5)(-3) + b, which simplifies to -1 = 27/5 + b.
Solving for b gives b = -32/5.
So, the equation of the line tangent to the curve at the point (-3,-1) is y = (-9/5)x - 32/5.