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 Partial Fractions: 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 1 - 2013 - Paper 9

Question 1

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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 the... 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 1 - 2013 - Paper 9

Step 1

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

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Answer

To solve for ( \frac{dy}{dx} ), we need to differentiate the equation implicitly with respect to x.

Starting from the equation: $x^2 + 4xy + y^2 + 27 = 0$

Differentiating gives:

For ( x^2 ): ( 2x )
For ( 4xy ): ( 4(x \cdot \frac{dy}{dx} + y) ) using the product rule
For ( y^2 ): ( 2y \cdot \frac{dy}{dx} )
The derivative of the constant (27) is 0.

Setting up the equation from the derivatives: $2x + 4y + 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

Now, isolating ( \frac{dy}{dx} ):

Combine like terms: $4x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - 4y$

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

Finally, we find: $\frac{dy}{dx} = \frac{-2x - 4y}{4x + 2y}$

Step 2

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

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Answer

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

$4x + 2y = 0$

From this, we express y in terms of x: $y = -2x$

Next, substitute ( y = -2x ) back into the original curve equation: $x^2 + 4x(-2x) + (-2x)^2 + 27 = 0$

Simplifying: $x^2 - 8x^2 + 4x^2 + 27 = 0$ $-2x^2 + 27 = 0$

Solving for x gives: $2x^2 = 27 \\ x^2 = \frac{27}{2} \\ x = -\sqrt{\frac{27}{2}}$ (considering negative x)

Therefore, the x-coordinate of Q is: $x = -\frac{3\sqrt{3}}{2}$

Next, we find the corresponding y-coordinate using ( y = -2x ): $y = -2 \left(-\frac{3\sqrt{3}}{2}\right) = 3\sqrt{3}$

So, the coordinates of Q are: $Q \left(-\frac{3\sqrt{3}}{2}, 3\sqrt{3}\right)$

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 1 - 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 1 - 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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