Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2 = 25(x^2-y^2) at the point ( -3 , -1 )
differentiate implicitly with respect to x and solve for dy/dx
sub in the point (-3,-1) to get the slope
now use the grade 9 way of finding the equation of a straight line
To find the equation of the tangent line to the given curve at a specific point, you can follow these steps:
1. Differentiate the equation of the curve with respect to x to find the derivative.
2. Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line.
3. Use the point-slope form of a line, y = mx + b, and substitute the slope and the coordinates of the point into the equation.
4. Solve for b, the y-intercept of the tangent line.
5. Form the equation of the tangent line using the slope and y-intercept.
Let's go through these steps in detail:
Step 1:
The equation of the curve is 2(x^2 + y^2)^2 = 25(x^2 - y^2).
Differentiating both sides of this equation with respect to x using the chain rule:
d/dx [2(x^2 + y^2)^2] = d/dx [25(x^2 - y^2)]
Using the chain rule, we get:
4(x^2 + y^2)(2x + 2y(dy/dx)) = 25(2x - 2y(dy/dx))
Step 2:
Substituting the given point (-3, -1) into the derivative, we get:
4((-3)^2 + (-1)^2)(2(-3) + 2(-1)(dy/dx)) = 25(2(-3) - 2(-1)(dy/dx))
Simplifying the equation:
4(10)(-8(dy/dx)) = 25(-4 - 2(dy/dx))
-320(dy/dx) = -100(-4 - 2(dy/dx))
-320(dy/dx) = 400 + 200(dy/dx)
(520)(dy/dx) = 400
dy/dx = 400/520
dy/dx = 10/13
Step 3:
Using the point-slope form of a line, y = mx + b, substitute the slope dy/dx = 10/13 and the coordinates of the point (-3, -1):
y - (-1) = (10/13)(x - (-3))
y + 1 = (10/13)(x + 3)
Step 4:
To find the y-intercept, solve the equation for y = 0:
0 + 1 = (10/13)(x + 3)
(10/13)(x + 3) = 1
10(x + 3) = 13
10x + 30 = 13
10x = 13 - 30
10x = -17
x = -17/10
The y-intercept is when x = -17/10, and the equation becomes:
y + 1 = (10/13)(-17/10 + 3)
y + 1 = (10/13)(-17 + 30)/10
y + 1 = (10/13)(13)/10
y + 1 = 13/13
y + 1 = 1
y = 1 - 1
y = 0
So, the y-intercept is 0.
Step 5:
Using the slope (10/13) and y-intercept (b = 0), we can form the final equation of the tangent line:
y = (10/13)x + 0
Simplifying the equation:
y = (10/13)x
Therefore, the equation of the tangent line to the curve 2(x^2 + y^2)^2 = 25(x^2 - y^2) at the point (-3, -1) is y = (10/13)x.