A curve is described by the equation $$x^2 + 4xy + y^2 + 27 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y. A point Q lies on the curve. The tangent to the curve at Q is parallel to the y-axis. Given that the x coordinate of Q is negative, (b) use your answer to part (a) to find the coordinates of Q.

A-Level Edexcel Maths Pure Basic Trigonometry: A curve is described by the equa

A curve is described by the equation $$x^2 + 4xy + y^2 + 27 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 9

Question 8

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A curve is described by the equation $$x^2 + 4xy + y^2 + 27 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y. A point Q lies on the curve. The tangent to t... show full transcript

Worked Solution & Example Answer:A curve is described by the equation $$x^2 + 4xy + y^2 + 27 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of x and y - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 9

Step 1

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

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Answer

To find ( \frac{dy}{dx} ), we start by differentiating the equation implicitly with respect to x:

$2x + 4y + 4x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$

Rearranging gives:

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

Factoring out ( \frac{dy}{dx} ):

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

Thus, we have:

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

Step 2

use your answer to part (a) to find the coordinates of Q.

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Answer

Since the tangent at Q is parallel to the y-axis, ( \frac{dy}{dx} ) must be undefined. This occurs when the denominator equals zero:

$4x + 2y = 0 \Rightarrow y = -2x$

Substituting (y = -2x) into the original equation:

$x^2 + 4x(-2x) + (-2x)^2 + 27 = 0$

This simplifies to:

$x^2 - 8x^2 + 4x^2 + 27 = 0 \Rightarrow -3x^2 + 27 = 0 \Rightarrow 3x^2 = 27 \Rightarrow x^2 = 9 \Rightarrow x = -3 \text{ (since x is negative)}$

Now substitute back to find y:

$y = -2(-3) = 6$

Therefore, the coordinates of Q are ((-3, 6)).

A curve is described by the equation $$x^2 + 4xy + y^2 + 27 = 0$$ (a) Find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 9

Worked Solution & Example Answer:A curve is described by the equation $$x^2 + 4xy + y^2 + 27 = 0$$ (a) Find \( \frac{dy}{dx} \) in terms of x and y - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 9

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

use your answer to part (a) to find the coordinates of Q.

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