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Determine The Equations Of The Lines Tangent To The Curve 3x3y2+2x2y=Y+4x Whenever X=1 .

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Final answer:

To find the equations of the lines tangent to the given curve, we need to determine the derivative of the curve and substitute the value of X=1. The resulting equation will give us the slope of the tangent lines. By substituting the values of X=1, dy/dx, and the slope into the straight line equation formula, we can determine the equations of the tangent lines.

Step-by-step explanation:

To find the equations of the lines tangent to the curve, we need to determine the derivative of the given curve. Taking the derivative of the curve equation, we get:

9x^2y^2 + 6xy^2 + 6x^2y - 1 = dy/dx + 4

Next, we substitute X=1 into the derived equation to find the slope of the tangent line:

9y^2 + 6y^2 + 6y - 1 = dy/dx + 4

Simplifying, we have:

15y^2 + 6y - 5 = dy/dx

Now, we substitute X=1 and dy/dx into the straight line equation formula y = mx + c to find the equations of the tangent lines:

For the first tangent line:

y = (15y^2 + 6y - 5)x + (3y - 7)

For the second tangent line:

y = (15y^2 + 6y - 5)x + (3y - 5)

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