Find an equation of the tangent line to the graph at the given point? 3(x² + y²)² = 100(x² + y²), (-4, -2)

Solution:

Given, the equation is 3(x² + y²)² = 100(x² + y²)

We have to find the equation of the tangent line to the graph at the point (-4, -2).

On expanding,

3(x⁴ + y⁴ + 2x²y²) = 100x² + 100y²

3x⁴ + 3y⁴ + 6x²y² = 100x² + 100y²

On differentiating,

12x³ + 12y³(dy/dx) + 6(2xy² + 2x²y(dy/dx)) = 200x + 200y(dy/dx)

12x³ + 12y³(dy/dx) + 12xy² + 12x²y(dy/dx) = 200x + 200y(dy/dx)

By grouping,

12y³(dy/dx) + 12x²y(dy/dx) - 200y(dy/dx) = 200x -12x³ - 12xy²

dy/dx(12y³ + 12x²y - 200y) = 200x - 12x³ - 12xy²

dy/dx = (200x - 12x³ - 12xy²)/(12y³ + 12x²y - 200y)

At the point (-4, -2),

200x - 12x³ - 12xy² = 200(-4) - 12(-4)³ - 12(-4)(-2)²

= -800 - 12(-64) - 12(-16)

= -800 + 768 + 192

= -800 + 960

= 160

12y³ + 12x²y - 200y = 12(-2)³ + 12(-4)²(-2) - 200(-2)

= 12(-8) + 12(-32) + 400

= -96 - 384 + 400

= -80

dy/dx = 160/(-80)

dy/dx = -2

The equation of the line in slope point form is given by

(y - y₁) = m(x - x₁)

(y - (-2)) = -2(x - (-4))

(y + 2) = -2(x + 4)

y + 2 = -2x - 8

y = -2x - 8 - 2

y = -2x - 10

Therefore, the equation of the line is y = -2x - 10.

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Find an equation of the tangent line to the graph at the given point? 3(x² + y²)² = 100(x² + y²), (-4, -2)

Summary:

An equation of the tangent line to the graph 3(x² + y²)² = 100(x² + y²) at the given point (-4, -2) is y = -2x - 10.