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.
The given ellipse is
\[ \frac{x^2}{4}+\frac{y^2}{9}=1 \] Since \(9\gt 4\), the major axis is along the \(y\)-axis.
So,
\[ a^2=9,\qquad b^2=4 \] \[ a=3,\qquad b=2 \]

Step 2: Find the value of \(c\).
For an ellipse,
\[ c^2=a^2-b^2 \] Therefore,
\[ c^2=9-4=5 \] \[ c=\sqrt{5} \]

Step 3: Write the coordinates of the foci.
Since the major axis is along the \(y\)-axis, the foci are
\[ (0,\pm c) \] Hence, the foci are
\[ F_1=(0,\sqrt{5}),\qquad F_2=(0,-\sqrt{5}) \]

Step 4: Find the distance from the point to \(F_1\).
Given point is
\[ P=\left(\frac{4}{\sqrt{5}},\frac{3}{\sqrt{5}}\right) \] Now,
\[ PF_1=\sqrt{\left(\frac{4}{\sqrt{5}}-0\right)^2+\left(\frac{3}{\sqrt{5}}-\sqrt{5}\right)^2} \] \[ =\sqrt{\frac{16}{5}+\left(\frac{3-5}{\sqrt{5}}\right)^2} \] \[ =\sqrt{\frac{16}{5}+\frac{4}{5}} \] \[ =\sqrt{\frac{20}{5}} \] \[ =2 \]

Step 5: Find the distance from the point to \(F_2\).
\[ PF_2=\sqrt{\left(\frac{4}{\sqrt{5}}-0\right)^2+\left(\frac{3}{\sqrt{5}}+\sqrt{5}\right)^2} \] \[ =\sqrt{\frac{16}{5}+\left(\frac{3+5}{\sqrt{5}}\right)^2} \] \[ =\sqrt{\frac{16}{5}+\frac{64}{5}} \] \[ =\sqrt{\frac{80}{5}} \] \[ =4 \]

Step 6: Final conclusion.
Therefore, the focal distances are
\[ 4,\;2 \] Hence,
\[ \boxed{4,2} \]