 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv-resit/2/a/solution

F

D

not containing

F

e > 1

is the locus

{P : dist (P, F) = e dist (P, D)} .

Put

c = a^{2} + b^{2}

and

e = c / a > 1

. For

F = (c, 0)

and

D = {x = a^{2} / c}

, the defining equality becomes

(x - c)^{2} + y^{2} = \frac{c}{a} x - \frac{a ^{2}}{c} .

Both sides are nonnegative, so squaring is reversible. Expanding and cancelling the linear terms gives

x^{2} - 2 c x + c^{2} + y^{2} = \frac{c ^{2}}{a ^{2}} x^{2} - 2 c x + a^{2} ⟺ \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1.

Thus the two matched focus of a conic and directrix of a conic choices, related by reflection, are

F_{\pm} = (\pm a^{2} + b^{2}, 0), D_{\pm} : x = \pm \frac{a ^{2}}{a ^{2} + b ^{2}}, e = 1 + \frac{b ^{2}}{a ^{2}} .

The signs in each pair must agree.

y = \pm (b / a) x^{2} - a^{2}

∣ x ∣ \to \infty

∣ x ∣ - x^{2} - a^{2} = \frac{a ^{2}}{∣ x ∣ + x ^{2} - a ^{2}} ⟶ 0.

Hence the perpendicular distance to one of the two straight lines tends to zero on each end of a branch. The asymptotes are therefore

y = \frac{b}{a} x and y = - \frac{b}{a} x .

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