Problem 87 - Algebra and Trigonometry

Question

Find the standard form of the equation of an ellipse with vertices at \((0,-6)\) and \((0,6),\) passing through \((2,-4)\)

Step-by-Step Solution

Answer

The standard equation of the ellipse is \( \frac{y^2}{36} + \frac{x^2}{28} = 1\).

Step 1: Determining the Center and 'a' value

Since the provided vertices are \((0,-6)\) and \((0,6)\), the center of the ellipse is at the origin \((0,0)\). 'a' is the distance from the center to either vertex, thus \(a = 6\). This gives the part of our equation: \((\frac{y^2}{6^2})\).

Step 2: Finding the 'b' value

The given point \((2,-4)\) lies on the ellipse. Substituting this point into the equation \((\frac{x^2}{a^2}) + (\frac{y^2}{b^2}) = 1\), keeping 'a' as 6 and solve for 'b', we get \((\frac{2^2}{b^2}) + (\frac{-4^2}{6^2}) = 1\). Solving the equation gives \(b = 2\sqrt{7}\).

Step 3: Forming the Standard Equation

We substitute 'a' and 'b' into the general form of the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), giving us the final standard form: \( \frac{y^2}{36} + \frac{x^2}{28} = 1\).