Equation of the line tangent to the graph of x^2-3xy=10 at point (1,-3)

Answer 1

The derivative of a function gives the slope of the tangent line, so we can the the derivative to find "m" in the "y = mx + b) equation of a line.

Since x² - 3xy = 10 is an equation in both x and y, it would be easier if we take the derivative implicitly. So, d(x² - 3xy)/dx = d(10)/dx. The derivative of x² with respect to x is easy, it's just 2x. The derivative of -3xy is tougher because we must use the product rule. It would be -3[x'y + xy') or -3[y + x(dy/dx)] or -3y -3x(dy/dx). The derivative of 10 with respect to x is 0 so putting it together we get: 2x - 3y - 3x(dy/dx) = 0 or dy/dx = (2x - 3y)/3x. Since we know the point on the curve is (1, -3), we know that x = 1 and y = -3 so dy/dx = [2(1) - 3(-3)]/[3(1)] = 11/3

Now we can use the point-slope format [(y - y₁) = m(x - x₁)] using m = 11/3 and the point as (1, -3) so the equation of the tangent line is y + 3 = 11/3(x - 1).

If you want, you can switch this around to the slope intercept form by multiplying it out and combining like-terms to get y = 11/3x -20/3