find an equation of the tanegent line to the curve 3xsquared + 5ysquared=8 at the point (1,1)

To find the equation of the tangent line to a curve at a given point, you need to find the slope of the curve at that point.

First, let's find the derivative of the curve equation with respect to x.

Differentiating the equation 3x^2 + 5y^2 = 8 with respect to x, we get:
6x + 10y * dy/dx = 0

Now, let's solve for dy/dx:
dy/dx = -6x / (10y)

To find the slope of the curve at the point (1,1), substitute x = 1 and y = 1 into the equation:
dy/dx = -6(1) / (10(1))
dy/dx = -6/10
dy/dx = -3/5

Now that we have the slope (-3/5), we can use the point-slope form of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

Substituting x1 = 1, y1 = 1, and m = -3/5, the equation becomes:
y - 1 = (-3/5)(x - 1)

Simplifying further:
y - 1 = (-3/5)x + 3/5

Bringing all terms to one side:
(-3/5)x + y - 1 - 3/5 = 0

Combining like terms:
(-3/5)x + y - 8/5 = 0

Therefore, the equation of the tangent line to the curve 3x^2 + 5y^2 = 8 at the point (1,1) is:
(-3/5)x + y - 8/5 = 0