Given:
- Area (\(A\)) of the ellipse = 50 square units
- Difference between semi-major and semi-minor axis lengths = 4 units
We'll use the formula \(A = \pi a b\) to solve for \(a\) and \(b\).
Step 1: Express \(a\) in terms of \(b\)
We know that \(a = b + 4\).
Step 2: Substitute into the Area Formula
Substitute \(a = b + 4\) into the area formula:
\[ 50 = \pi (b + 4) b \]
Step 3: Solve for \(b\)
\[ 50 = \pi (b^2 + 4b) \]
\[ 50 = \pi b^2 + 4\pi b \]
\[ \pi b^2 + 4\pi b - 50 = 0 \]
This is a quadratic equation in terms of \(b\). We can solve it using the quadratic formula:
\[ b = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = \pi \)
- \( b = 4\pi \)
- \( c = -50 \)
Plugging in the values, we get:
\[ b = \frac{-4\pi \pm \sqrt{(4\pi)^2 - 4(\pi)(-50)}}{2\pi} \]
\[ b = \frac{-4\pi \pm \sqrt{16\pi^2 + 200\pi}}{2\pi} \]
\[ b = \frac{-4\pi \pm \sqrt{16\pi^2 + 200\pi}}{2\pi} \]
\[ b = \frac{-4 \pm \sqrt{16 + 200}}{2} \]
\[ b = \frac{-4 \pm \sqrt{216}}{2} \]
\[ b = \frac{-4 \pm 6\sqrt{6}}{2} \]
\[ b = -2 \pm 3\sqrt{6} \]
Since the length cannot be negative, we take the positive value:
\[ b = -2 + 3\sqrt{6} \]
\[ b \approx 7.9 \]
Now, we can calculate \(a\):
\[ a = b + 4 \]
\[ a = 7.9 + 4 \]
\[ a = 11.9 \]
So, the semi-major axis length (\(a\)) is approximately 11.9 units and the semi-minor axis length (\(b\)) is approximately 7.9 units.
