find the equation to the tangent line of the parabola given by x^2=2y at point (3, 9/2)
y = x^2/2
y' = x
so, the line through (3,9/2) with slope=3 is
y - 9/2 = 3(x-3)
To find the equation of the tangent line to the parabola given by x^2 = 2y at the point (3, 9/2), we can use the concept of derivatives.
Step 1: Find the derivative of the equation x^2 = 2y with respect to x. The derivative of x^2 with respect to x is 2x, and the derivative of 2y with respect to x is 2(dy/dx).
Therefore, the derivative of the equation x^2 = 2y is 2x = 2(dy/dx).
Step 2: At the given point (3, 9/2), substitute x = 3 into the derivative equation to find the slope of the tangent line (dy/dx).
2(3) = 2(dy/dx)
6 = 2(dy/dx)
(dy/dx) = 6/2
(dy/dx) = 3
Step 3: Now that we have the slope of the tangent line, we can use the point-slope form of the equation of a line to find the equation.
The point-slope form of a line is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) represents the point on the line, and m represents the slope of the line.
Substituting the values into the equation, we have:
y - 9/2 = 3(x - 3)
Simplifying the equation, we get:
y - 9/2 = 3x - 9
Rearranging the equation, we get:
y = 3x - 9 + 9/2
Simplifying further, we get:
y = 3x - 9/2
Therefore, the equation of the tangent line to the parabola x^2 = 2y at the point (3, 9/2) is y = 3x - 9/2.