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 Integration: 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 3 - 2018 - Paper 9

Question 3

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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 coord... 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 3 - 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 perform implicit differentiation on the equation

$x^2 + xy + y^2 - 4x - 5y + 1 = 0,$
we differentiate each term with respect to x:

For (x^2), the derivative is (2x).
For (xy), we use the product rule: (\frac{d}{dx}(xy) = x\frac{dy}{dx} + y).
For (y^2), the derivative is (2y\frac{dy}{dx}).
For (-4x), the derivative is (-4).
For (-5y), the derivative is (-5\frac{dy}{dx}).
The derivative of a constant (1) is 0.

After differentiation, the equation is:

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

Now, we can rearrange this equation to isolate (\frac{dy}{dx}):

$x\frac{dy}{dx} + 2y\frac{dy}{dx} - 5\frac{dy}{dx} = 4 - 2x - y.$
Factoring out (\frac{dy}{dx}) gives us:

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

Thus, we find:

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

Step 2

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

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Answer

Setting (\frac{dy}{dx} = 0), we have:

$\frac{4 - 2x - y}{x + 2y - 5} = 0.$
This implies the numerator must equal zero:

$4 - 2x - y = 0,$
which rearranges to:

$y = 4 - 2x.$

Now we substitute this expression for y back into the original curve equation:

$x^2 + x(4 - 2x) + (4 - 2x)^2 - 4x - 5(4 - 2x) + 1 = 0.$
This simplifies to:

$x^2 + 4x - 2x^2 + 16 - 16x + 4x^2 - 16 + 10x - 5 + 1 = 0.$
Combining like terms gives:

$2x^2 - 2x + 0 = 0.$
Factoring out 2 we get:

$x^2 - x = 0$
which can be factored to:

$x(x - 1) = 0.$
Thus, the solutions are:

$x = 0 \text{ and } x = 1.$
Therefore, the x coordinates of the two points where ( \frac{dy}{dx} = 0 ) are (0) and (1).

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 3 - 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 3 - 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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