Let and be the foci of the ellipse . Point moves on the ellipse. Find the minimum value of

Analytic Geometry Solution – Problem 20: Find the minimum value of

Question

Let $M(2,0)$ and $N(-2,0)$ be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ . Point $P$ moves on the ellipse. Find the minimum value of $\frac{9}{|PM|}+\frac{1}{|PN|}.$

Step-by-step solution

For any point $P$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ , the sum of distances to the foci is constant: $|PM|+|PN|=2a=2\cdot 3=6.$ Apply Cauchy–Schwarz: $\frac{9}{|PM|}+\frac{1}{|PN|}=\frac{3^2}{|PM|}+\frac{1^2}{|PN|}\ge \frac{(3+1)^2}{|PM|+|PN|}=\frac{16}{6}=\frac{8}{3}.$ Equality holds when $|PM|:|PN|=3:1$ . Therefore the minimum value is $\frac{8}{3}$ .

Final answer

$\min\left(\frac{9}{|PM|}+\frac{1}{|PN|}\right)=\frac{8}{3}.$

Marking scheme

1. Checkpoints (max 4 pts total)

Ellipse distance sum (1 pt)

State $|PM|+|PN|=2a=6$ . (1 pt)

Inequality + equality (3 pts)

Apply Cauchy–Schwarz (or equivalent AM-GM) correctly to obtain $\ge \frac{16}{|PM|+|PN|}$ . (2 pts)
Substitute $|PM|+|PN|=6$ and conclude minimum $\frac{8}{3}$ ; mention equality condition. (1 pt)

Total (max 4)

2. Zero-credit items

Treating $|PM|$ and $|PN|$ as independent without using $|PM|+|PN|$ constant.
Guessing the minimizing ratio without an inequality argument.

3. Deductions

C-S setup error (-1): using $\frac{9}{|PM|}+\frac{1}{|PN|}\ge \frac{10^2}{|PM|+|PN|}$ (wrong weights).
Constant-sum error (-1): using $|PM|+|PN|=10$ instead of 6.

Analytic Geometry – Problem 20: Find the minimum value of