The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y. (b) Find the x coordinates of the two points on C where $ \frac{dy}{dx} = 0 $ Give exact answers in their simplest form. (Solutions based entirely on graphical or numerical methods are not acceptable.)

A-Level Edexcel Maths Pure Differentiation: The curve C has equation $$x^2

The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 9

Question 4

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The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y. (b) Find the x coo... show full transcript

Worked Solution & Example Answer:The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 9

Step 1

Use implicit differentiation to find $ \frac{dy}{dx} $ in terms of x and y.

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Answer

To differentiate the given equation implicitly, we first take the derivative of both sides with respect to x:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) - \frac{d}{dx}(4x) - \frac{d}{dx}(5y) + \frac{d}{dx}(1) = 0$

Calculating each term gives us:

$2x + \left(\frac{dy}{dx} \cdot x + y\right) + 2y\frac{dy}{dx} - 4 - 5\frac{dy}{dx} = 0$

Rearranging and collecting ( \frac{dy}{dx} ) terms:

$\left(x + 2y - 5\right)\frac{dy}{dx} = 4 - 2x - y$

Thus,

$\frac{dy}{dx} = \frac{4 - 2x - y}{x + 2y - 5}$

Step 2

Find the x coordinates of the two points on C where $ \frac{dy}{dx} = 0 $

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Answer

For ( \frac{dy}{dx} = 0 ), the numerator must equal zero:

$4 - 2x - y = 0$

This implies:

$y = 4 - 2x$

Next, we substitute ( y ) back into the original equation:

$x^2 + x(4 - 2x) + (4 - 2x)^2 - 4x - 5(4 - 2x) + 1 = 0$

Expanding and simplifying:

$x^2 + 4x - 2x^2 + 16 - 16x + 4x^2 - 20 + 10x + 1 = 0$

Combine like terms:

$3x^2 - 2x - 3 = 0$

Using the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-3)}}{2\cdot 3} = \frac{2 \pm \sqrt{4 + 36}}{6} = \frac{2 \pm \sqrt{40}}{6}$

Simplifying gives:

$x = \frac{1 \pm \sqrt{10}}{3}$

Thus, the x-coordinates where ( \frac{dy}{dx} = 0 ) are:

$x = \frac{1 + \sqrt{10}}{3}, \quad x = \frac{1 - \sqrt{10}}{3}$

The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 9

Worked Solution & Example Answer:The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 9

Use implicit differentiation to find \( \frac{dy}{dx} \) in terms of x and y.

Find the x coordinates of the two points on C where \( \frac{dy}{dx} = 0 \)

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