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 Modelling with Functions: 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 2 - 2013 - Paper 9

Question 2

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

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 ta... 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 2 - 2013 - Paper 9

Step 1

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

96%

114 rated

Answer

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

Differentiate the equation: [ x^2 + 4xy + y^2 + 27 = 0 ] Applying implicit differentiation: [ 2x + 4\left( y + x\frac{dy}{dx} \right) + 2y\frac{dy}{dx} = 0 ] Simplifying this gives: [ 2x + 4y + 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 ]
Rearranging the equation to isolate ( \frac{dy}{dx} ): [ 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - 4y ] [ (4x + 2y)\frac{dy}{dx} = -2x - 4y ] [ \frac{dy}{dx} = \frac{-2x - 4y}{4x + 2y} ] Thus, ( \frac{dy}{dx} ) is: [ \frac{dy}{dx} = \frac{-2(x + 2y)}{2(2x + y)} = \frac{-(x + 2y)}{2x + y} ]

Step 2

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

99%

104 rated

Answer

From part (a), we have:

Condition for tangent to be parallel to the y-axis implies ( \frac{dy}{dx} = 0 ): Setting ( \frac{-(x + 2y)}{2x + y} = 0 ) leads to: [ x + 2y = 0 \quad (1) ] Therefore, ( y = -\frac{x}{2} ).
Substituting into original equation: Now, substitute ( y = -\frac{x}{2} ) into the original curve equation: [ x^2 + 4x\left(-\frac{x}{2}\right) + \left(-\frac{x}{2}\right)^2 + 27 = 0 ] Simplifying this gives: [ x^2 - 2x^2 + \frac{x^2}{4} + 27 = 0 ] Combine terms: [ -x^2 + \frac{x^2}{4} + 27 = 0 ] Multiply through by 4 to eliminate the fraction: [ -4x^2 + x^2 + 108 = 0 ] [ -3x^2 + 108 = 0 ] [ 3x^2 = 108 ] [ x^2 = 36 ] Thus, ( x = -6 ) (since the x coordinate of Q is negative).
Finding y-coordinate: Substitute ( x = -6 ) back into equation (1): [ y = -\frac{-6}{2} = 3 ]
Coordinates of Q: Therefore, the coordinates of Q are ( (-6, 3) ).

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 2 - 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 2 - 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

Join the A-Level students using SimpleStudy...