Photo AI
The curve C has the equation $2x + 3y^2 + 3x^2y = 4x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 8
Question 3
The curve C has the equation $2x + 3y^2 + 3x^2y = 4x^2$. The point P on the curve has coordinates $(-1, 1)$. (a) Find the gradient of the curve at P. (b) Hence fin... show full transcript
Worked Solution & Example Answer:The curve C has the equation $2x + 3y^2 + 3x^2y = 4x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 8
Step 1
Find the gradient of the curve at P.
Answer
To find the gradient of the curve at point P, we first need to differentiate the equation implicitly with respect to .
Starting from the equation:
Differentiating both sides: rac{d}{dx}(2x) + rac{d}{dx}(3y^2) + rac{d}{dx}(3x^2y) = rac{d}{dx}(4x^2)
From which we get: 2 + 6y rac{dy}{dx} + (3x^2 rac{dy}{dx} + 6xy) = 8x
Now, substituting and to find the gradient: 2 + 6(1) rac{dy}{dx} + (3(-1)^2 rac{dy}{dx} + 6(-1)(1) = 8(-1) 2 + 6 rac{dy}{dx} + 3 rac{dy}{dx} - 6 = -8 9 rac{dy}{dx} - 4 = -8 9 rac{dy}{dx} = -4 rac{dy}{dx} = - rac{4}{9}
Thus, the gradient of the curve at P is - rac{4}{9}.
Step 2
Hence find the equation of the normal to C at P.
Answer
The gradient of the normal line is the negative reciprocal of the gradient of the curve.
Given that the gradient of the curve at P is - rac{4}{9}, the gradient of the normal line, , is: m_n = - rac{1}{(- rac{4}{9})} = rac{9}{4}
Now, using the point-slope form of a line to find the equation of the normal line: Substituting and m_n = rac{9}{4}: y - 1 = rac{9}{4}(x + 1)
Expanding this: y - 1 = rac{9}{4}x + rac{9}{4} y = rac{9}{4}x + rac{13}{4}
To express this in the form , rearranging gives: - rac{9}{4}x + y - rac{13}{4} = 0
Multiplying through by 4 to eliminate the fractions results in:
Thus, the equation of the normal line at P is: