Q: If x2 +xy + y2 = 3, what is the equation of the tangent line to this graph when y = 1.
A: To find the equation of a tangent line, we know our final answer will be in y=mx+b format. To get this, we will need to find a point and a slope.
Step 1: Establish our point of interest. When y = 1, we need to know what x equals. Plug 1 into the equation for y to get:
x2 +x*(1) + (1)2 = 3
x2 +x – 2 = 0
(x + 2)(x – 1) = 0
x = -2, 1.
So, we have two points of interest: (x, y) = (-2, 1) and (1, 1). We will have 2 different tangent lines when y = 1.
Step 2: Finding the slope(s): To find the slope at one point, we need a derivative!
Take the derivative of this equation implicitly as follows (let y’ be the same as dy/dx):
d/dx(x2 +xy + y2 = 3) –> 2x + y + xy’ + 2yy’ = 0
Note: Implicit differentiation above required a product rule and a chain rule. We are taking the derivative of each piece with respect to the variable x!
Now, solve for y’ (since y’ is the slope):
2x + y + xy’ + 2yy’ = 0
xy’ + 2yy’ = -2x – y
y'(x + 2y) = -2x – y
y’ = (-2x – y) / (x + 2y)
Now, plug in each point separately to find the slope at the given point:
Point One, (-2, 1), yields:
y’ = undefined (dividing by zero). The slope is undefined at this point, so there is no tangent line.
Point Two, (1, 1), yields:
y’ = -1.
So, the slope at the point (1, 1) is -1
Step Three: The equation of the tangent line:
y = mx + b
Plug in our point and slope to solve for b!
1 = -1*1 + b
1 = -1 + b
2 = b.
The equation of the tangent line at the point (1, 1) to the given graph is y = -x+2!