Question

The equation of conic section whose vertices are \((0, \pm 8)\) and asymptotes are \(y = \pm \frac{4}{3}x\) is given by

• A. \(\frac{x^2}{36} + \frac{y^2}{64} = 1\)
• B. \(\frac{x^2}{36} - \frac{y^2}{64} = 1\)
• C. \(\frac{x^2}{64} + \frac{y^2}{36} = 1\)
• D. \(-\frac{x^2}{36} + \frac{y^2}{64} = 1\)

Hint:

For any hyperbola, look at the vertices first:
- Vertices on the x-axis (\( \pm a, 0 \)) \(\rightarrow\) Horizontal hyperbola (\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)).
- Vertices on the y-axis (\( 0, \pm a \)) \(\rightarrow\) Vertical hyperbola (\( -\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \)).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question requires finding the equation of a conic section (specifically, a hyperbola) given its vertices and the equations of its asymptotes.

Step 2: Detailed Explanation:
Let us analyze the given geometric features:
- Vertices: The vertices are given as \( (0, \pm 8) \).
Since the vertices lie on the y-axis (the x-coordinate is 0), the transverse axis of the hyperbola is vertical.
This means we are dealing with a vertical hyperbola centered at the origin, which has the standard equation:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \quad \text{or} \quad -\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \] The coordinates of the vertices of a vertical hyperbola are \( (0, \pm a) \).
Comparing this with the given vertices \( (0, \pm 8) \), we find:
\[ a = 8 \implies a^2 = 64 \] - Asymptotes: The equations of the asymptotes for a vertical hyperbola are given by the formula:
\[ y = \pm \frac{a}{b} x \] We are given the asymptotes as:
\[ y = \pm \frac{4}{3} x \] Comparing the coefficients:
\[ \frac{a}{b} = \frac{4}{3} \] Substitute the value of \( a = 8 \) into this ratio:
\[ \frac{8}{b} = \frac{4}{3} \] Solve for \( b \):
\[ 4b = 24 \implies b = 6 \implies b^2 = 36 \] - Formulate the Equation:
Substitute the values of \( a^2 = 64 \) and \( b^2 = 36 \) back into the standard vertical hyperbola equation:
\[ -\frac{x^2}{36} + \frac{y^2}{64} = 1 \] This matches the expression in Option (D).

Step 3: Final Answer:
The equation of the conic section is \( -\frac{x^2}{36} + \frac{y^2}{64} = 1 \).
Therefore, the correct choice is Option (D).