A-Level Edexcel Maths Pure Rational Expressions: A curve C has the equation $$x^

A curve C has the equation $$x^3 + 2xy - x - y^3 - 20 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y: (b) Find an equation of the tangent to C at the point (3, –2), giving your answer in the form $ ax + by + c = 0 $, where a, b and c are integers. - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 7

Question 3

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A curve C has the equation $$x^3 + 2xy - x - y^3 - 20 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y: (b) Find an equation of the tangent to C at the poin... show full transcript

Worked Solution & Example Answer:A curve C has the equation $$x^3 + 2xy - x - y^3 - 20 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y: (b) Find an equation of the tangent to C at the point (3, –2), giving your answer in the form $ ax + by + c = 0 $, where a, b and c are integers. - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 7

Step 1

Find $ \frac{dy}{dx} $ in terms of x and y:

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Answer

To find ( \frac{dy}{dx} ), we will differentiate the given equation implicitly with respect to x:

Starting with: $x^3 + 2xy - x - y^3 - 20 = 0$

Differentiate each term:
- The derivative of ( x^3 ) is ( 3x^2 ).
- The derivative of ( 2xy ) requires the product rule: ( 2 \left( x \frac{dy}{dx} + y \right) ).
- The derivative of ( -x ) is ( -1 ).
- The derivative of ( -y^3 ) is ( -3y^2 \frac{dy}{dx} ).
- The derivative of ( -20 ) is 0.
Putting it all together: $3x^2 + 2 \left( x \frac{dy}{dx} + y \right) - 1 - 3y^2 \frac{dy}{dx} = 0$
Rearranging gives: $3x^2 + 2y - 1 + \left( 2x - 3y^2 \right) \frac{dy}{dx} = 0$
Isolating ( \frac{dy}{dx} ): $\left( 2x - 3y^2 \right) \frac{dy}{dx} = -3x^2 - 2y + 1$
Finally, solving for ( \frac{dy}{dx} ): $\frac{dy}{dx} = \frac{-3x^2 - 2y + 1}{2x - 3y^2}$

Step 2

Find an equation of the tangent to C at the point (3, –2), giving your answer in the form $ ax + by + c = 0 $, where a, b and c are integers:

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Answer

First, we substitute the point (3, -2) into the derivative found in part (a): $\frac{dy}{dx} = \frac{-3(3)^2 - 2(-2) + 1}{2(3) - 3(-2)^2}$
Calculating this: $\frac{dy}{dx} = \frac{-27 + 4 + 1}{6 - 18} = \frac{-22}{-12} = \frac{11}{6}$
The slope of the tangent line at the point (3, -2) is ( \frac{11}{6} ).
Using the point-slope form of the equation of a line: ( y - y_1 = m(x - x_1) ), we have: $y - (-2) = \frac{11}{6}(x - 3)$
Simplifying this: $y + 2 = \frac{11}{6}x - \frac{33}{6}$ $y = \frac{11}{6}x - \frac{33}{6} - 2$ $y = \frac{11}{6}x - \frac{45}{6}$
Rewriting in the form ( ax + by + c = 0 ): $\frac{11}{6}x - y - \frac{45}{6} = 0$ Multiplying through by 6 to eliminate the fraction: $11x - 6y - 45 = 0$
Thus, the values of a, b, and c are 11, -6, and -45 respectively.

Find \( \frac{dy}{dx} \) in terms of x and y:

Find an equation of the tangent to C at the point (3, –2), giving your answer in the form \( ax + by + c = 0 \), where a, b and c are integers:

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