 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/10/solution

18 x + 32 y \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x}_{(1, 1)} = - \frac{9}{16} .

The tangent line through

(1, 1)

is therefore

y - 1 = - \frac{9}{16} (x - 1), equivalently 9 x + 16 y = 25.

(9, 16)

, parallel to the gradient

(18, 32)

9 x^{2} + 16 y^{2}

at that point.

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