The graph of 3 left parenthesis x squared plus y squared right parenthesis squared equals 100 left parenthesis x squared minus y squared right parenthesis 3(x^2+y^2)2=100(x^2−y^2). Find the equation of the tangent line at the point left parenthesis 4 comma 2 right parenthesis(4,2).
The Correct Answer and Explanation is :
To find the equation of the tangent line to the given curve ( 3(x^2 + y^2)^2 = 100(x^2 – y^2) ) at the point ((4, 2)), follow these steps:
1. Differentiate the given equation implicitly
Start with:
[
3(x^2 + y^2)^2 = 100(x^2 – y^2)
]
Differentiate both sides with respect to (x):
[
6(x^2 + y^2)(2x + 2y\frac{dy}{dx}) = 100(2x – 2y\frac{dy}{dx})
]
Simplify:
[
12(x^2 + y^2)(x + y\frac{dy}{dx}) = 200x – 200y\frac{dy}{dx}
]
Distribute and reorganize terms:
[
12x(x^2 + y^2) + 12y(x^2 + y^2)\frac{dy}{dx} = 200x – 200y\frac{dy}{dx}
]
Group terms involving (\frac{dy}{dx}) on one side:
[
12y(x^2 + y^2)\frac{dy}{dx} + 200y\frac{dy}{dx} = 200x – 12x(x^2 + y^2)
]
Factor out (\frac{dy}{dx}):
[
\frac{dy}{dx}\left[12y(x^2 + y^2) + 200y\right] = 200x – 12x(x^2 + y^2)
]
Solve for (\frac{dy}{dx}):
[
\frac{dy}{dx} = \frac{200x – 12x(x^2 + y^2)}{12y(x^2 + y^2) + 200y}
]
2. Substitute the point ((4, 2))
First, compute (x^2 + y^2) at ((4, 2)):
[
x^2 + y^2 = 4^2 + 2^2 = 16 + 4 = 20
]
Now substitute (x = 4) and (y = 2) into the formula for (\frac{dy}{dx}):
[
\frac{dy}{dx} = \frac{200(4) – 12(4)(20)}{12(2)(20) + 200(2)}
]
Simplify:
[
\frac{dy}{dx} = \frac{800 – 960}{480 + 400} = \frac{-160}{880} = -\frac{2}{11}
]
3. Find the tangent line equation
The slope of the tangent line is (-\frac{2}{11}), and the point is ((4, 2)). Use the point-slope form of a line:
[
y – y_1 = m(x – x_1)
]
Substitute (m = -\frac{2}{11}), (x_1 = 4), and (y_1 = 2):
[
y – 2 = -\frac{2}{11}(x – 4)
]
Simplify:
[
y = -\frac{2}{11}x + \frac{8}{11} + 2
]
Combine terms:
[
y = -\frac{2}{11}x + \frac{30}{11}
]
Final Answer:
The equation of the tangent line is:
[
y = -\frac{2}{11}x + \frac{30}{11}
]
Explanation (300 Words):
The equation (3(x^2 + y^2)^2 = 100(x^2 – y^2)) describes a curve involving both (x) and (y) in a complex relation. To find the tangent line, we determine the slope at the specific point ((4, 2)) using implicit differentiation because (y) is a function of (x), but not explicitly given.
Implicit differentiation is applied to both sides of the equation. Using the chain rule carefully handles terms like ((x^2 + y^2)^2), and derivative terms of (y) produce (\frac{dy}{dx}). Grouping terms and solving for (\frac{dy}{dx}) isolates the slope of the tangent line.
At ((4, 2)), substituting (x = 4) and (y = 2) into the differentiated equation gives (\frac{dy}{dx} = -\frac{2}{11}). The tangent line is then formed using the point-slope formula (y – y_1 = m(x – x_1)), where (m) is the slope, and ((x_1, y_1)) is the given point. This results in (y = -\frac{2}{11}x + \frac{30}{11}).
The process highlights key calculus principles: applying implicit differentiation, handling higher-degree equations, and ensuring correct substitution for exact results.