Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x^2+2xy-y^2+x=6 , (2,4) (hyperbola)
f'(x) =
2x +2y + 2xdy/dx - 2ydy/dx + 1 = 0
2xdy/dx - 2ydy/dx = -2x - 2y -1
dy/dx = (-2x - 2y - 1)/ (2x - 2y)
the tangent passes through (2,4)
m = dy/dx = (-2(2) - 2(4) -1 )
------------------
(2(2) - 2(4))
m = 13/4
the equation of the tangent line through (2,4);
y = mx + c
4 = 13/4(2) + c
c = -5/2
y = 13/4x - 5/2
thanks for the help.
To find the equation of the tangent line to the curve at the given point, we will first differentiate the equation implicitly with respect to x, and then substitute the values of the given point into the resulting equation.
Let's begin by differentiating the given equation with respect to x. When we differentiate with respect to x, we treat y as a function of x and use the chain rule. Recall that the chain rule states that if y is a function of x, then d/dx(y) = dy/dx.
Differentiating x^2 + 2xy - y^2 + x = 6 with respect to x, we get:
2x + 2y(dy/dx) - 2y(dy/dx) - 2y + 1 = 0
Simplifying this equation, we have:
2x + 1 - 2y = 2y(dy/dx)
Now, let's determine the value of dy/dx at the point (2, 4) by substituting x = 2 and y = 4 into the equation. Rearranging the equation, we get:
2x + 1 - 2y = 2y(dy/dx)
2(2) + 1 - 2(4) = 2(4)(dy/dx)
4 + 1 - 8 = 8(dy/dx)
-3 = 8(dy/dx)
Hence, dy/dx = -3/8.
Now that we have found dy/dx, we can use the point-slope form of a line to write the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope, which in this case is dy/dx.
So, substituting the values of (x1, y1) = (2, 4) and m = -3/8, we have:
y - 4 = (-3/8)(x - 2)
Simplifying this equation, we get the equation of the tangent line:
y - 4 = (-3/8)x + 3/4
or
y = (-3/8)x + 3/4 + 4
Final equation of the tangent line:
y = (-3/8)x + 19/4