How do I find the equation of the line tangent to the graph of the equation x² + 3xy – 10 at the point (1, 3)?

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To find the equation of the line tangent to the graph of the equation x² + 3xy – 10 = 0 at the point (1, 3), we need to follow these steps:

  1. Differentiate the equation implicitly: Since the equation is not solved for y, we will use implicit differentiation. Starting with the equation:

    2. x² + 3xy – 10 = 0

    Differentiating both sides with respect to x gives:

    2x + 3(y + x(dy/dx)) = 0

    Here, dy/dx represents the derivative of y with respect to x. We can rewrite the equation:

    2x + 3y + 3x(dy/dx) = 0
  2. Isolate dy/dx: Rearrange the equation to isolate dy/dx:
    3. 3x(dy/dx) = -2x - 3y
    dy/dx = (-2x - 3y) / (3x)
  3. Evaluate dy/dx at the point (1, 3): Now, substitute (x, y) = (1, 3) into the derivative:
    4. dy/dx = (-2(1) - 3(3)) / (3(1)) = (-2 - 9) / 3 = -11 / 3
  4. Find the equation of the tangent line: We now have the slope of the tangent line at the point (1, 3), which is -11/3. Using the point-slope form of the equation of a line, we can express the equation of the tangent line:
    5. y - y1 = m(x - x1)

    Substituting in the values:

    y - 3 = (-11/3)(x - 1)
  5. Rearranging the equation: Distributing and rearranging, we’ll get:
    6. y - 3 = -11/3 x + 11/3
    y = -11/3 x + 11/3 + 3
    y = -11/3 x + 20/3

Thus, the equation of the tangent line to the graph of the equation x² + 3xy – 10 at the point (1, 3) is:

y = -11/3 x + 20/3.

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