 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/4/solution

X = x^{2}

and

Y = y^{2}

in the two conic sections. A real intersection requires

X, Y \geq 0

. Solving the resulting linear equations gives

X = \frac{3}{2 a + 1}, Y = \frac{a - 1}{2 a + 1} .

For

a > 0

X

are positive, so intersections exist exactly when

a \geq 1

. At

a = 1

they are

(\pm 1, 0)

; for

a > 1

both choices of sign for each square root give intersections. Consequently

0 < a < 1 gives no shared points .

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