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C4 Differentiation

A curve has equation x^2 + 2xy – 3y^2 + 16 = 0. Find the coordinates of the points on the curve where dy/dx = 0.

There was no mark scheme for this, so please could some tell me if they also got,

(2, -2) (-2, 2)

Thanks,

Varun

Reply 1

AlphaNumeric

I'm guessing xydd is supposed to be

\frac{dy}{dx}

x^{2}+2xy-3y^{2}+16=0

Differentiate with respect to x

2x+2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0

(2x-6y)\frac{dy}{dx} + 2x+2y = 0

\frac{dy}{dx} = \frac{-2x-2y}{2x-6y}

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

\frac{dy}{dx} = 0 \Rightarrow x+y=0 \Rightarrow x=-y

Put this back into the original equation gives

x^{2}+2x(-x)-3(-x)^{2}+16=0

x^{2}-2x^{2}-3x^{2}+16=0

4x^{2}=16

x = \pm 2

Therefore the points are (-2,2) and (2,-2)

Reply 2

samd

Is this on of the earlier questions from C4 June 2005 ?

Reply 3

7589200

Yea it is dy/dx. Thanks very much.

yea, it is. Did you do this? :smile:

Reply 4

the bear

AlphaNumeric

\frac{dy}{dx} = \frac{-2x-2y}{2x-6y}

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

can you spot the mistake in the lower line ?

Reply 5

damaster275

the bear

can you spot the mistake in the lower line ?

I can, but it doesn't affect the answer anyway...

Reply 6

Rockitty99

Yes thank you so much for this, I was looking for this answer.

Reply 7

Notnek

Original post by the bear

can you spot the mistake in the lower line ?

PROM 2005 @the bear :smile:

Reply 8

the bear

Original post by Notnek

PROM 2005 @the bear :smile:

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