Find the tangent line to the ellipse defined by 3x^2 + 5y^2 = 17 at the point (-2, 1).
6 x dx/dx + 10 y dy/dx = 0
6 (-2) + 10 (1) dy/dx = 0
10 * slope = 12
slope = 1.2
y = 1.2 x + b
1 = 1.2(-2) + b
solve for b
To find the tangent line to the ellipse at the point (-2, 1), we need to find the derivative of the ellipse function with respect to x and evaluate it at the given point.
Let's start by rearranging the equation of the ellipse to isolate y:
3x^2 + 5y^2 = 17
5y^2 = 17 - 3x^2
y^2 = (17 - 3x^2) / 5
Next, take the square root of both sides:
y = ±√((17 - 3x^2) / 5)
Since we are looking for the tangent line at the point (-2, 1), we substitute x = -2 and find the corresponding y-coordinate:
y = ±√((17 - 3(-2)^2) / 5)
y = ±√((17 - 3(4)) / 5)
y = ±√((17 - 12) / 5)
y = ±√(5 / 5)
y = ±1
So, the given point (-2, 1) lies on the ellipse. Now, let's find the derivative of the ellipse function with respect to x:
dy/dx = d(±√((17 - 3x^2) / 5))/dx
To make things simpler, we can just focus on finding the derivative of the square root term.
Using the chain rule, the derivative of √((17 - 3x^2) / 5) with respect to x is:
1/(2√((17 - 3x^2) / 5)) * d((17 - 3x^2) / 5)/dx
1/(2√((17 - 3x^2) / 5)) * (-6x/5)
Now, we can substitute x = -2 into the derived expression:
dy/dx = 1/(2√((17 - 3(-2)^2) / 5)) * (-6(-2)/5)
dy/dx = 1/(2√((17 - 3(4)) / 5)) * (12/5)
dy/dx = 1/(2√((17 - 12) / 5)) * (12/5)
dy/dx = 1/(2√(5/5)) * (12/5)
dy/dx = 1/(2√1) * (12/5)
dy/dx = 1/2 * (12/5)
dy/dx = 6/5
The derivative dy/dx is equal to 6/5 at the point (-2, 1).
Now, we have the slope of the tangent line, which is the derivative dy/dx evaluated at the given point. Using the point-slope form of a line, we can write the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is the point (-2, 1) and m is the slope 6/5:
y - 1 = (6/5)(x - (-2))
y - 1 = (6/5)(x + 2)
y - 1 = (6/5)x + 12/5)
y = (6/5)x + 12/5 + 1
y = (6/5)x + 12/5 + 5/5
y = (6/5)x + 17/5
Therefore, the equation of the tangent line to the ellipse 3x^2 + 5y^2 = 17 at the point (-2, 1) is y = (6/5)x + 17/5.