Find the equation of tangent to the curve at a point $(-2,-1)$ given $2 x^2 y^2-4 x y^3-x^2 y^2+3 e^{2 x} y+8 x=0$

Simplify the given equation of the curve: x 2 y 2 − 4 x y 3 + 3 e 2 x y + 8 x = 0 .
Use implicit differentiation to find d x d y ​ = 2 x 2 y − 12 x y 2 + 3 e 2 x − 2 x y 2 + 4 y 3 − 6 y e 2 x − 8 ​ .
Evaluate d x d y ​ at the point ( − 2 , − 1 ) to find the slope m ≈ − 0.4914 .
Use the point-slope form to find the tangent line equation: y = − 0.4914 x − 1.9829 . The equation of the tangent line is y = − 0.4914 x − 1.9829 ​ .

Explanation

Problem Setup We are given the equation of a curve: 2 x 2 y 2 − 4 x y 3 − x 2 y 2 + 3 e 2 x y + 8 x = 0 . We want to find the equation of the tangent line to this curve at the point ( − 2 , − 1 ) .

Simplify the Equation First, simplify the equation: 2 x 2 y 2 − 4 x y 3 − x 2 y 2 + 3 e 2 x y + 8 x = 0 becomes x 2 y 2 − 4 x y 3 + 3 e 2 x y + 8 x = 0 .

Implicit Differentiation Next, we need to find d x d y ​ using implicit differentiation. Differentiating both sides of the equation x 2 y 2 − 4 x y 3 + 3 e 2 x y + 8 x = 0 with respect to x , we get:


d x d ​ ( x 2 y 2 ) − d x d ​ ( 4 x y 3 ) + d x d ​ ( 3 e 2 x y ) + d x d ​ ( 8 x ) = 0
Applying the product rule and chain rule, we have:
( 2 x y 2 + x 2 ( 2 y d x d y ​ )) − ( 4 y 3 + 4 x ( 3 y 2 d x d y ​ )) + ( 3 e 2 x d x d y ​ + 3 y ( 2 e 2 x )) + 8 = 0
2 x y 2 + 2 x 2 y d x d y ​ − 4 y 3 − 12 x y 2 d x d y ​ + 3 e 2 x d x d y ​ + 6 y e 2 x + 8 = 0

Solve for dy/dx Now, we solve for d x d y ​ :

d x d y ​ ( 2 x 2 y − 12 x y 2 + 3 e 2 x ) = − 2 x y 2 + 4 y 3 − 6 y e 2 x − 8
d x d y ​ = 2 x 2 y − 12 x y 2 + 3 e 2 x − 2 x y 2 + 4 y 3 − 6 y e 2 x − 8 ​

Evaluate dy/dx at (-2, -1) We are given the point ( − 2 , − 1 ) . Substitute x = − 2 and y = − 1 into the expression for d x d y ​ :

d x d y ​ = 2 ( − 2 ) 2 ( − 1 ) − 12 ( − 2 ) ( − 1 ) 2 + 3 e 2 ( − 2 ) − 2 ( − 2 ) ( − 1 ) 2 + 4 ( − 1 ) 3 − 6 ( − 1 ) e 2 ( − 2 ) − 8 ​
d x d y ​ = − 8 − 24 + 3 e − 4 − 4 − 4 + 6 e − 4 − 8 ​ = − 32 + 3 e − 4 − 16 + 6 e − 4 ​ = − 16 + 2 3 ​ e − 4 − 8 + 3 e − 4 ​ = 16 − 2 3 ​ e − 4 8 − 3 e − 4 ​
d x d y ​ = 2 ( 16 − 2 3 ​ e − 4 ) 2 ( 8 − 3 e − 4 ) ​ = 32 − 3 e − 4 16 − 6 e − 4 ​
d x d y ​ = − 16 + 3 e − 4 − 8 + 3 e − 4 ​
Using a calculator, we find that d x d y ​ ≈ − 16 + 3 ( 0.0183 ) − 8 + 3 ( 0.0183 ) ​ ≈ − 16 + 0.0549 − 8 + 0.0549 ​ ≈ − 15.9451 − 7.9451 ​ ≈ 0.4983

Calculate dy/dx using tool Using the python tool, we found that d x d y ​ = 3 e − 4 + 16 − 8 + 6 e − 4 ​ ≈ − 0.4914439273839698 .

Find the Tangent Line Equation Now we use the point-slope form of a line to find the equation of the tangent line: y − y 1 ​ = m ( x − x 1 ​ ) , where ( x 1 ​ , y 1 ​ ) = ( − 2 , − 1 ) and m = − 0.4914439273839698 .


y − ( − 1 ) = − 0.4914439273839698 ( x − ( − 2 ))
y + 1 = − 0.4914439273839698 ( x + 2 )
y = − 0.4914439273839698 x − 0.9828878547679396 − 1
y = − 0.4914439273839698 x − 1.9828878547679396

Final Answer Therefore, the equation of the tangent line is approximately y = − 0.4914 x − 1.9829 .

Final Answer The equation of the tangent line is approximately y = − 0.4914 x − 1.9829 .


Examples
In physics, understanding tangent lines is crucial for analyzing motion. For example, if you have a position function of an object over time, the tangent line at a specific time gives you the instantaneous velocity of the object at that moment. This concept extends to various fields, such as engineering, where tangent lines help determine rates of change and optimize designs.

Answered by GinnyAnswer | 2025-07-03

To find the tangent line to the curve at the point (-2, -1), we used implicit differentiation to find the slope, yielding d x d y ​ ≈ − 0.4914 . Then, we applied the point-slope form of the line equation, resulting in the tangent line equation y = − 0.4914 x − 1.9829 .
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Answered by Anonymous | 2025-07-04