A curve has equation $3x^2 - y^2 + xy = 4$. The points $P$ and $Q$ lie on the curve. The gradient of the tangent to the curve is $rac{3}{5}$ at $P$ and at $Q$. (a) Use implicit differentiation to show that $y - 2x = 0$ at $P$ and at $Q$. (b) Find the coordinates of $P$ and $Q$.

A-Level Edexcel Maths Pure General Binomial Expansion: A curve has equation $3x^2

A curve has equation $3x^2 - y^2 + xy = 4$ - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 7

Question 5

A curve has equation $3x^2 - y^2 + xy = 4$. The points $P$ and $Q$ lie on the curve. The gradient of the tangent to the curve is $rac{3}{5}$ at $P$ and at $Q$. (... show full transcript

Worked Solution & Example Answer:A curve has equation $3x^2 - y^2 + xy = 4$ - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 7

Step 1

Use implicit differentiation to show that $y - 2x = 0$ at $P$ and at $Q$

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Answer

To differentiate the equation $3x^2 - y^2 + xy = 4$ implicitly with respect to $x$ , we apply the derivatives to each term:

The derivative of $3x^2$ is $6x$ .
The derivative of $-y^2$ is $-2y rac{dy}{dx}$ .
The derivative of $xy$ $x y$ uses the product rule:
- Here, the derivative is $x rac{dy}{dx} + y$ .

Setting the derivatives equal results in:

$6x - 2y rac{dy}{dx} + rac{dy}{dx} (x) + y = 0$

Rearranging gives:

$6x + y - 2y rac{dy}{dx} + x rac{dy}{dx} = 0$

Next, isolate $rac{dy}{dx}$ :

$rac{dy}{dx} (x - 2y) = - (6x + y)$

Thus, we have:

$rac{dy}{dx} = rac{- (6x + y)}{x - 2y}$

Setting $rac{dy}{dx} = rac{3}{5}$ :

$rac{3}{5} = rac{-(6x + y)}{x - 2y}$

Cross-multiplying leads to:

$3(x - 2y) = -5(6x + y)$

Expanding and combining terms:

$3x - 6y + 30x + 5y = 0$

This simplifies to:

$33x - y = 0$

Hence, at points $P$ and $Q$ , we find that $y - 2x = 0$ .

Step 2

Find the coordinates of $P$ and $Q$

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Answer

To find the coordinates of points $P$ and $Q$ , we substitute $y = 2x$ into the original equation:

Substituting into the curve's equation:

$3x^2 - (2x)^2 + x(2x) = 4$

This simplifies to:

$3x^2 - 4x^2 + 2x^2 = 4$

Combining the terms yields:

$x^2 = 4$

Thus:

$x = 2 ext{ or } x = -2$

Now substituting these values back to find $y$ :

For $x = 2$ : $y = 2(2) = 4 ext{, so } P(2, 4)$
For $x = -2$ : $y = 2(-2) = -4 ext{, so } Q(-2, -4)$

Thus, the coordinates are:

Points $P(2, 4)$ and $Q(-2, -4)$ .

A curve has equation $3x^2 - y^2 + xy = 4$ - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 7

Worked Solution & Example Answer:A curve has equation $3x^2 - y^2 + xy = 4$ - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 7

Use implicit differentiation to show that $y - 2x = 0$ at $P$ and at $Q$

Find the coordinates of $P$ and $Q$

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