The curve C has equation
Find an equation for the tangent to C at the point P (2, 3), giving your answer in the form ax+ by + c = 0 where a, b and c are integers.
(6)
- Implicit differentiation
- Point-slope form
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Differentiate Both Sides with Respect to x
Given:
\( xy^2 = x^2y + 6 \)
Apply the product rule to both sides:Left Side (\( xy^2 \)):
\( \frac{d}{dx} (xy^2) = y^2 + x \cdot 2y \frac{dy}{dx} \)
(Product rule: \(\frac{d}{dx} (uv) = u’v + uv’,\) where \(u = x, v = y^2\))
Right Side (\( x^2y + 6 \)):
\( \frac{d}{dx} (x^2y) + \frac{d}{dx} (6) = 2xy + x^2 \frac{dy}{dx} \)
(Product rule: \(u = x^2, v = y\))
Combined Result:
\( y^2 + 2xy \frac{dy}{dx} = 2xy + x^2 \frac{dy}{dx} \)
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Solve for \(\frac{dy}{dx}\)
Group terms containing \(\frac{dy}{dx}\):
\( 2xy \frac{dy}{dx} – x^2 \frac{dy}{dx} = 2xy – y^2 \)
Factor out \(\frac{dy}{dx}\):
\( \frac{dy}{dx} (2xy – x^2) = 2xy – y^2 \)
Isolate \(\frac{dy}{dx}\):
\( \frac{dy}{dx} = \frac{2xy – y^2}{2xy – x^2} \)
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Evaluate at \(P(2,3)\)
Substitute \(x = 2\), \(y = 3\):
\( \frac{dy}{dx} = \frac{2(2)(3) – 3^2}{2(2)(3) – 2^2} = \frac{12 – 9}{12 – 4} = \frac{3}{8} \)
Hence, the gradient of the tangent at \(P(2,3)\) is \(\frac{3}{8}\). -
Use Point-Slope Form
Equation of a line:
\( y – y_1 = m(x – x_1) \)
Substitute \(m = \frac{3}{8}\) and \(P(2,3)\):
\( y – 3 = \frac{3}{8}(x – 2) \)
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Convert to Standard Form (\(ax + by + c = 0\))
Multiply through by 8 to eliminate fractions:
\( 8(y – 3) = 3(x – 2) \)
Expand:
\( 8y – 24 = 3x – 6 \)
Rearrange:
\( 3x – 8y + 18 = 0 \)