Question

The eccentricity of the ellipse represented by the equation \(7x^2+16y^2-14x+64y-377 = 0\) is...

• A. \(\frac{3}{4}\)
• B. \(\frac{\sqrt{7}}{4}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{3}{8}\)

Hint:

Complete the squares to reach standard form, then use e squared = 1 - b squared / a squared.

Answer 1

The Correct Option is A

Step 1: Complete the square
Group the terms: \(7(x^2 - 2x) + 16(y^2 + 4y) = 377\). Add the squares: \(7(x-1)^2 - 7 + 16(y+2)^2 - 64 = 377\), so
\[ 7(x-1)^2 + 16(y+2)^2 = 448 \]

Step 2: Standard form
Divide by 448: \[ \frac{(x-1)^2}{64} + \frac{(y+2)^2}{28} = 1 \]
Here \(a^2 = 64\) and \(b^2 = 28\), with the major axis along the x-direction because 64 is bigger.

Step 3: Find e
\[ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{28}{64} = \frac{36}{64} \Rightarrow e = \frac{6}{8} = \frac{3}{4} \]

Step 4: Check the options
\(\frac{\sqrt{7}}{4}\) equals \(\sqrt{1 - \frac{9}{16}}\), which is the value of \(b/a\) for this ellipse, a common mix-up. Options (C) and (D) do not come from the numbers 64 and 28.

Final Answer:
The eccentricity is 3/4. This is option (A). \[ \boxed{\text{(A) }\frac{3}{4}} \]