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 Incoming links: Past exam of the mathematical methods I course of the University of Oxford / 2021 / 10 / a

Past exam of the mathematical methods I course of the University of Oxford / 2021 / 10 / b / Solution 2026-10-25

The two real slope roots of

p^{2} + f p - 1 = 0

are distinct because their discriminant is

f^{2} + 4 > 0

. Vieta's formulas give

p_{+} p_{-} = - 1

, so the tangent vectors

(1, p_{+})

and

(1, p_{-})

have zero scalar product. Whenever the coefficients define local solution curves, the two families are therefore orthogonal trajectories.

For the stated example at

(3, 4)

, the slopes are

3

and

- 1/3

. In the family from part a, the constants are

1

and

- 1/9

y = \frac{x ^{2} - 1}{2}, y = \frac{9}{2} - \frac{x ^{2}}{18} .

Both curves pass through

(3, 4)

, and

3 (- 1/3) = - 1

verifies the right angle.

Figure 1.
Two members of the same parabolic family pass through (3,4) with perpendicular tangent lines
. Details.

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