find the slope of the tangent line at the point (1,2) for the ellipse 2x^2+3y^2=14
y'(1) = ??
differentiate implicitly
4x + 6y dy/dx = 0
dy/dx = -6y/(4x) = -3y/(2x)
sub in (1,2) to get the slope
take it from there
To find the slope of the tangent line at the point (1, 2) for the ellipse 2x^2 + 3y^2 = 14, we need to find the derivative of y with respect to x and evaluate it at x = 1.
Step 1: Rewrite the given equation in terms of y:
2x^2 + 3y^2 = 14
3y^2 = 14 - 2x^2
y^2 = (14 - 2x^2)/3
Step 2: Differentiate both sides of the equation implicitly with respect to x:
d/dx(y^2) = d/dx[(14 - 2x^2)/3]
2y * dy/dx = -4x/3
Step 3: Solve for dy/dx, which represents the derivative of y with respect to x:
dy/dx = (-4x/3) / 2y
dy/dx = -2x / (3y)
Step 4: Substitute the coordinates of the point (1, 2) into the derived equation to find the slope:
dy/dx = -2(1) / (3(2))
dy/dx = -1/3
Therefore, the slope of the tangent line at the point (1, 2) for the ellipse 2x^2 + 3y^2 = 14 is -1/3.