The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of $ x $ and $ y $. (b) Find the coordinates of the points on C where $ \frac{dy}{dx} = 0 $ (Solutions based entirely on graphical or numerical methods are not acceptable.)

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

The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of $ x $ and $ y $ - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 4

Question 3

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The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of $ x $ and $ y $. (b) Find the coordinates of the points on C w... show full transcript

Worked Solution & Example Answer:The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find $ \frac{dy}{dx} $ in terms of $ x $ and $ y $ - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 4

Step 1

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

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Answer

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

Starting from the equation:

$x^2 - 3xy - 4y^2 + 64 = 0$

Differentiating each term gives:

Differentiating ( x^2 ) gives ( 2x )
Differentiating ( -3xy ) using the product rule: ( -3 \left( x \frac{dy}{dx} + y \right) )
Differentiating ( -4y^2 ) gives ( -8y \frac{dy}{dx} )
The derivative of ( 64 ) is ( 0 ).

Thus, we have:

$2x - 3 \left( x \frac{dy}{dx} + y \right) - 8y \frac{dy}{dx} = 0$

Rearranging and combining like terms leads to:

$2x - 3y - \left(3x + 8y\right) \frac{dy}{dx} = 0$

Solving for ( \frac{dy}{dx} ):

$\frac{dy}{dx} = \frac{2x - 3y}{3x + 8y}$

Step 2

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

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Answer

From the previous part, we find that ( \frac{dy}{dx} = 0 ) when the numerator is zero:

$2x - 3y = 0\Rightarrow y = \frac{2}{3}x$

Substituting ( y = \frac{2}{3}x ) into the original curve equation gives:

$x^2 - 3x\left(\frac{2}{3}x\right) - 4\left(\frac{2}{3}x\right)^2 + 64 = 0$

Simplifying this:

$x^2 - 2x^2 - \frac{16}{9}x^2 + 64 = 0$

Combining and finding a common denominator:

$\left(1 - 2 - \frac{16}{9}\right)x^2 + 64 = 0$

This results in:

$\left(-\frac{25}{9}\right)x^2 + 64 = 0$

Solving for ( x^2 ):

$\frac{25}{9} x^2 = 64 \Rightarrow x^2 = \frac{64 \times 9}{25} \Rightarrow x = \pm \frac{24}{5}$

Using ( y = \frac{2}{3}x ) to find corresponding ( y ) values:

When ( x = \frac{24}{5} ): ( y = \frac{2}{3} \left(\frac{24}{5}\right) = \frac{16}{5}) When ( x = -\frac{24}{5} ): ( y = \frac{2}{3} \left(-\frac{24}{5}\right) = -\frac{16}{5})

Thus the coordinates are:\n1. ( \left(\frac{24}{5}, \frac{16}{5}\right) )\n2. ( \left(-\frac{24}{5}, -\frac{16}{5}\right) )

The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find \( \frac{dy}{dx} \) in terms of \( x \) and \( y \) - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 4

Worked Solution & Example Answer:The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find \( \frac{dy}{dx} \) in terms of \( x \) and \( y \) - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 4

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

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

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