Question

If the ends of the major axis \(A'\) and \(A\) of the ellipse \[ \frac{(x-2)^2}{a^2}+\frac{(y-3)^2}{b^2}=1 \] are respectively at distances \(9\) and \(3\) units from a directrix \(L\), then the foci of the ellipse are:

• A. \(\left(2\pm\frac34,\;3\right)\)
• B. \(\left(2\pm\frac32,\;3\right)\)
• C. \(\left(\pm\frac32,\;3\right)\)
• D. \(\left(\pm\frac34,\;3\right)\)

Hint:

For an ellipse, \[ c=ae, \qquad c^2=a^2-b^2. \] Once the eccentricity \(e\) is known, finding the foci usually becomes a straightforward substitution problem.

Answer 1

The Correct Option is B

Step 1: Understand the picture.
We have an ellipse with centre $(2,3)$. The two ends of the major axis are at distances $9$ and $3$ from one directrix. From these we recover $a$ and the eccentricity, then locate the foci.

Step 2: Use the directrix distances.
The two vertices lie on the major axis, one nearer and one farther from the directrix, at distances $3$ and $9$. The centre, being midway, is at distance $\dfrac{3+9}{2}=6$ from the directrix. The half-length is $a=\dfrac{9-3}{2}=3$.

Step 3: Find the eccentricity.
The directrix is at distance $\dfrac{a}{e}$ from the centre. So $\dfrac{a}{e}=6$, giving $\dfrac{3}{e}=6$, hence $e=\dfrac{1}{2}$.

Step 4: Find $c$, the focal distance from centre.
Since $c=ae$, \[ c=3\times\frac{1}{2}=\frac{3}{2}. \]

Step 5: Locate the foci.
The major axis is horizontal (the $x$-term carries $a^2$), so the foci are at $(2\pm c,\,3)$.

Step 6: Write the answer.
\[ \left(2\pm\frac{3}{2},\;3\right). \] This is option 2.
\[ \boxed{\left(2\pm\dfrac{3}{2},\;3\right)} \]