Find the focal distance a and first-quadrant intersection Q of the ellipse and the circle whose diameter joins its foci.

01 / Standard#Your turn

\frac{x^2}9+\frac{y^2}4=1,\quad F_1=(-a,0),\ F_2=(a,0),\ a>0

Official paper · jm02-2026 · 3(a) · PDF 5

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Subtract the squared semiaxes to find a².

Hint 2

The focal circle is centred at the origin.

Worked solution

Find the focus and the circle equation.
$a^2=9-4=5,\quad a=\sqrt5,\quad x^2+y^2=5$
Eliminate y² and choose positive coordinates.
$4x^2+9y^2=36,\quad5x^2=9,\quad Q=\left(\frac{3\sqrt5}5,\frac{4\sqrt5}5\right)$

a=√5; Q=(3√5/5,4√5/5).

Checks and common pitfalls: Here a names the focal distance in the question, not the major semiaxis.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Find the focus and the circle equation.
$a^2=9-4=5,\quad a=\sqrt5,\quad x^2+y^2=5$
Eliminate y² and choose positive coordinates.
$4x^2+9y^2=36,\quad5x^2=9,\quad Q=\left(\frac{3\sqrt5}5,\frac{4\sqrt5}5\right)$

Think first. Reveal a hint when the class is ready.

2026 JM02