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.

Answer 2

Step 1: Understand the problem and setup the ellipse equation.

We are given an ellipse whose major axis lies along the x-axis. The general equation of such an ellipse is:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \]

where \( a \) is the semi-major axis (along the x-axis) and \( b \) is the semi-minor axis (along the y-axis).

The ellipse passes through the points \( (4, 3) \) and \( (-1, 4) \). Substituting these points into the ellipse equation will allow us to find relationships between \( a^2 \) and \( b^2 \).

Step 2: Substitute \( (4, 3) \) into the ellipse equation.

Substitute \( x = 4 \) and \( y = 3 \):

\[ \frac{4^2}{a^2} + \frac{3^2}{b^2} = 1. \]

Simplify:

\[ \frac{16}{a^2} + \frac{9}{b^2} = 1. \tag{1} \]

Step 3: Substitute \( (-1, 4) \) into the ellipse equation.

Substitute \( x = -1 \) and \( y = 4 \):

\[ \frac{(-1)^2}{a^2} + \frac{4^2}{b^2} = 1. \]

Simplify:

\[ \frac{1}{a^2} + \frac{16}{b^2} = 1. \tag{2} \]

Step 4: Solve the system of equations.

We now have two equations:

\[ \frac{16}{a^2} + \frac{9}{b^2} = 1 \tag{1} \]

\[ \frac{1}{a^2} + \frac{16}{b^2} = 1. \tag{2} \]

Let \( \frac{1}{a^2} = u \) and \( \frac{1}{b^2} = v \). Then the equations become:

\[ 16u + 9v = 1 \tag{3} \]

\[ u + 16v = 1. \tag{4} \]

Multiply equation (4) by 16:

\[ 16u + 256v = 16. \tag{5} \]

Subtract equation (3) from equation (5):

\[ (16u + 256v) - (16u + 9v) = 16 - 1, \]

\[ 247v = 15. \]

Solve for \( v \):

\[ v = \frac{15}{247}. \]

Substitute \( v = \frac{15}{247} \) into equation (4):

\[ u + 16\left(\frac{15}{247}\right) = 1, \]

\[ u + \frac{240}{247} = 1. \]

\[ u = 1 - \frac{240}{247} = \frac{7}{247}. \]

Step 5: Compute \( \frac{1}{a^2} + \frac{16}{b^2} \).

Recall that \( \frac{1}{a^2} = u \) and \( \frac{1}{b^2} = v \). Thus:

\[ \frac{1}{a^2} + \frac{16}{b^2} = u + 16v. \]

Substitute \( u = \frac{7}{247} \) and \( v = \frac{15}{247} \):

\[ \frac{1}{a^2} + \frac{16}{b^2} = \frac{7}{247} + 16\left(\frac{15}{247}\right). \]

\[ \frac{1}{a^2} + \frac{16}{b^2} = \frac{7}{247} + \frac{240}{247}. \]

\[ \frac{1}{a^2} + \frac{16}{b^2} = \frac{247}{247} = 1. \]

Final Answer:

\( 1 \)