Find the locus of A when its two tangents to the ellipse are perpendicular, including vertical-tangent cases.

01 / Standard#Your turn

\frac{x^2}{9}+\frac{y^2}{4}=1,\quad A=(h,k)\text{ outside the ellipse}

Official paper · jm02-2023 · 3(b) · PDF 5

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

For two finite slopes use m₁m₂=−1.

Hint 2

Add the intersections of vertical and horizontal tangent lines.

Worked solution

The finite-slope case yields a circle.
$\frac{k^2-4}{h^2-9}=-1\implies h^2+k^2=13$
The missing cases are the intersections x=±3 and y=±2; all lie on the same circle.
$(h,k)=(\pm3,\pm2)$
Every point of this circle lies outside the ellipse, so no further exclusion is needed.
$h^2/9+k^2/4\ge(h^2+k^2)/9=13/9>1$

The full circle x²+y²=13.

Checks and common pitfalls: Omitting the vertical cases would leave four unjustified holes.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The finite-slope case yields a circle.
$\frac{k^2-4}{h^2-9}=-1\implies h^2+k^2=13$
The missing cases are the intersections x=±3 and y=±2; all lie on the same circle.
$(h,k)=(\pm3,\pm2)$
Every point of this circle lies outside the ellipse, so no further exclusion is needed.
$h^2/9+k^2/4\ge(h^2+k^2)/9=13/9>1$

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

2023 JM02