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The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 9
Question 4
The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of x and y. (b) Find the x coord... show full transcript
Worked Solution & Example Answer:The curve C has equation $$x^2 + xy + y^2 - 4x - 5y + 1 = 0$$ (a) Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 9
Step 1
Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of x and y.
Answer
To find (\frac{dy}{dx}), we differentiate each term of the equation with respect to x:
- Differentiate (x^2): (\frac{d}{dx}(x^2) = 2x)
- Differentiate (xy) using the product rule:
- (x \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y)
- Differentiate (y^2): (\frac{d}{dx}(y^2) = 2y \frac{dy}{dx})
- Differentiate (-4x): (-4)
- Differentiate (-5y): (-5 \frac{dy}{dx})
- The constant (1) differentiates to (0).
Now, combine these results into one equation:
[ 2x + \left( x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} - 4 - 5 \frac{dy}{dx} = 0 ]
Rearranging gives:
[ (x + 2y - 5) \frac{dy}{dx} = 4 - 2x - y ]
Thus, solving for (\frac{dy}{dx}):
[ \frac{dy}{dx} = \frac{4 - 2x - y}{x + 2y - 5} ]
Step 2
Find the x coordinates of the two points on C where \(\frac{dy}{dx} = 0\).
Answer
To find where (\frac{dy}{dx} = 0), set the numerator equal to zero:
[ 4 - 2x - y = 0 ]
This simplifies to:
[ y = 4 - 2x ]
Next, substitute this expression for y back into the original equation of the curve:
[ x^2 + x(4 - 2x) + (4 - 2x)^2 - 4x - 5(4 - 2x) + 1 = 0 ]
Expanding this gives:
[ x^2 + 4x - 2x^2 + 16 - 16x + 4x^2 - 4x - 20 + 10x + 1 = 0 ]
Combine like terms:
[ 3x^2 - 4x - 3 = 0 ]
Now, use the quadratic formula to find x:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
where (a = 3), (b = -4), and (c = -3):
[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 36}}{6} = \frac{4 \pm \sqrt{52}}{6} = \frac{4 \pm 2\sqrt{13}}{6} = \frac{2 \pm \sqrt{13}}{3} ]
Thus, the x-coordinates of the two points where (\frac{dy}{dx} = 0) are:
[ x = \frac{2 + \sqrt{13}}{3}, \quad x = \frac{2 - \sqrt{13}}{3} ]