Final answer:
To find the equation of the tangent line to the curve 9x² 16y² = 52, differentiate the equation with respect to x to find the slope of the tangent line at a given point on the curve. Substitute the x-coordinate of the point into the derivative equation to find the slope. Use the point-slope form of a line to write the equation of the tangent line.
Step-by-step explanation:
To find the equation of the tangent line to the curve 9x² 16y² = 52, we need to find the slope of the tangent line at a given point on the curve. The slope of the tangent line at any point on a curve is equal to the derivative of the curve at that point. First, we'll differentiate the equation with respect to x:
18x + 32y * (dy/dx) = 0
To find the slope at a specific point, we substitute the x-coordinate of that point into the equation above and solve for dy/dx. For example, if we want to find the slope at the point (2, -1), we substitute x = 2 into the equation:
18(2) + 32(-1) * (dy/dx) = 0
Simplifying the equation gives:
36 - 32(dy/dx) = 0
Now solve for dy/dx:
32(dy/dx) = 36
(dy/dx) = 36/32
(dy/dx) = 9/8
So, the slope of the tangent line at the point (2, -1) is 9/8. Now we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
Substituting the values, we get:
y - (-1) = (9/8)(x - 2)
Simplifying the equation gives:
y + 1 = (9/8)x - (9/4)
y = (9/8)x - (13/4)