Find the equation of the tangent line at the point on the graph of the equation y^2-xy-12=0, where x=1. The tangent line at (1,-3) is what?
To find the equation of the tangent line at the point (1,-3), we need to find the derivative of the equation y^2 - xy - 12 = 0 and evaluate it at x=1.
Taking the derivative implicitly with respect to x, we get:
2y * dy/dx - y - x * dy/dx = 0
Rearranging the terms, we have:
(2y - x) * dy/dx = y
Dividing both sides by (2y - x), we obtain:
dy/dx = y / (2y - x)
Next, substitute x = 1 and y = -3 into the derivative equation to find the slope of the tangent line at (1,-3).
dy/dx = (-3) / (2 * (-3) - 1)
= (-3) / (-6 - 1)
= -3 / -7
= 3/7
So, the slope of the tangent line at (1,-3) is 3/7.
Finally, we can use the point-slope form to find the equation of the tangent line:
y - y1 = m(x - x1)
y - (-3) = (3/7)(x - 1)
y + 3 = (3/7)x - 3/7
y = (3/7)x - 3/7 - 3
y = (3/7)x - 3/7 - 21/7
y = (3/7)x - 24/7
Therefore, the equation of the tangent line at the point (1,-3) is y = (3/7)x - 24/7.