Find the equation of the tangent line to the graph of
4y^2− xy − 3 = 0, at the point P=(1,1)
To find the equation of the tangent line to the graph of the given equation at the point P(1, 1), we can use the method of finding the derivative of the equation and then substituting the coordinates of the point P into the derivative.
Step 1: Find the derivative of the equation
Differentiate the given equation with respect to x:
d/dx (4y^2 − xy − 3) = d/dx (0)
8y(dy/dx) - y - x(dy/dx) = 0
Step 2: Substitute the coordinates of point P(1, 1)
Now, substitute x = 1 and y = 1 into the derived equation:
8(1)(dy/dx) - (1) - (1)(dy/dx) = 0
7(dy/dx) - 1 = 0
Step 3: Solve for dy/dx
We can now solve the equation for dy/dx, which will give us the slope of the tangent line at point P(1, 1):
7(dy/dx) = 1
dy/dx = 1/7
Step 4: Write the equation of the tangent line
We have the slope of the tangent line at point P(1, 1), which is 1/7. Now, we need to find the y-intercept (b) of the tangent line using the coordinates of point P. The equation of the tangent line can be written in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
Using the point-slope form of a line, we have:
y - 1 = (1/7)(x - 1)
y - 1 = (1/7)x - 1/7
y = (1/7)x + (6/7)
Therefore, the equation of the tangent line to the graph of 4y^2 − xy − 3 = 0 at the point P(1, 1) is y = (1/7)x + (6/7).