Question

Suppose a and b are lengths of major and minor axes of an ellipse that passes through the points (4,3) and (-1,4). If the major axis of the ellipse lies along the x-axis , then the value of \(\frac{1}{a^2}+\frac{16}{b^2}\) is

• A. 4
• B. 1
• C. 2
• D. \(\frac{1}{2}\)
• E. \(\frac{1}{4}\)

Answer 1

The Correct Option is B

Since the major axis is along the x-axis, the equation of the ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Substituting the given points:
For point (4, 3): \( \frac{16}{a^2} + \frac{9}{b^2} = 1 \)
For point (-1, 4): \( \frac{1}{a^2} + \frac{16}{b^2} = 1 \)

Adding the two equations:
\( \frac{17}{a^2} + \frac{25}{b^2} = 2 \)

From the second equation: \( \frac{1}{a^2} = 1 - \frac{16}{b^2} \). Substituting into the first equation:
\( 16(1 - \frac{16}{b^2}) + \frac{9}{b^2} = 1 \)
\( 16 - \frac{256}{b^2} + \frac{9}{b^2} = 1 \)
\( 15 = \frac{247}{b^2} \)
\( b^2 = \frac{247}{15} \)
\( \frac{1}{b^2} = \frac{15}{247} \)

\( \frac{1}{a^2} = 1 - 16(\frac{15}{247}) = 1 - \frac{240}{247} = \frac{7}{247} \)

Therefore \( \frac{1}{a^2} + \frac{16}{b^2} = \frac{7}{247} + 16(\frac{15}{247}) = \frac{7}{247} + \frac{240}{247} = \frac{247}{247} = 1 \).

Checking with the other equation: \( \frac{16}{a^2} + \frac{9}{b^2} = 16(\frac{7}{247}) + 9(\frac{15}{247}) = \frac{112}{247} + \frac{135}{247} = \frac{247}{247} = 1 \).

So the value of \( \frac{1}{a^2} + \frac{16}{b^2} \) is 1.