Question

The focal distances of the point \(\left(\dfrac{4}{\sqrt{5}},\dfrac{3}{\sqrt{5}}\right)\) on the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\) are

• A. \(\dfrac{10}{3},\dfrac{2}{3}\)
• B. \(3,1\)
• C. \(\dfrac{13}{3},\dfrac{5}{3}\)
• D. \(4,2\)

Hint:

For the ellipse \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\), where \(a\gt b\), the foci are \((0,\pm c)\) and \(c^2=a^2-b^2\).

Answer 1

The Correct Option is D

Step 1: Identify the ellipse parameters.
For $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$, since $9>4$, the major axis is vertical with $a^2=9$ and $b^2=4$, so $a=3$ and $b=2$.
Step 2: Find the eccentricity.
Using $b^2 = a^2(1-e^2)$: $4 = 9(1-e^2)$, so $1-e^2 = \frac{4}{9}$, giving $e^2 = \frac{5}{9}$ and $e = \frac{\sqrt5}{3}$.
Step 3: Recall the focal distance formula.
For a vertical-major ellipse, the focal distances of a point $(x_1,y_1)$ are $a \pm e\,y_1$.
Step 4: Plug in the $y$-coordinate.
Here $y_1 = \dfrac{3}{\sqrt5}$, so $e\,y_1 = \dfrac{\sqrt5}{3}\cdot\dfrac{3}{\sqrt5} = 1$.
Step 5: Compute the two distances.
The focal distances are $a + e y_1 = 3 + 1 = 4$ and $a - e y_1 = 3 - 1 = 2$.
Step 6: State the answer.
Hence the focal distances of the point are $4$ and $2$.
\[ \boxed{4,\,2} \]