Question

Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 25x²+4y² – 150x + 32y + 189 = 0

Accepted Answer

Start by grouping the x-terms and y-terms together and move the constant to the other side: $$25x^{2} - 150x$$ + $$4y^{2} + 32y = -189$. Factor out the coefficients of the squared terms from their respective groups: 25(x$$^{2}$$ - 6x) + 4(y$$^{2}$$ + 8y) = -189. $$Complete the square for each group inside the parentheses. For $$x^{2} - 6x$, take half of -6, square it, and add inside the parentheses. Do the same for y$$^{2}$$ + 8y. $$Remember to balance the equation by adding the equivalent values outside the parentheses multiplied by their coefficients. Rewrite the equation with the completed squares as perfect square trinomials: $$25(x - h$$)^{2}$$ + 4(y + k$$)^{2}$$ = C$$, where $$h$$ and $$k$$ are the values found from completing the square, and $$C is $$the new constant on the right side. Divide the entire equation by $$C to $$get the standard form of the ellipse: $$\frac{(x - h$$)^{2}$$}{a$$^{2}$$} + \frac{(y + k$$)^{2}$$}{b$$^{2}$$} = 1. $$Then identify $$a$$, $$b$$, and calculate the foci using $$c^{2}$$ = |$$a^{2}$$ - $$b^{2}$$|$$, where the foci are located at (h$$ \pm c, k)$$ or (h$$, k \pm c)$$ depending on the major axis.

Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 25x²+4y² – 150x + 32y + 189 = 0

Verified video answer for a similar problem:

Key Concepts

Completing the Square

Standard Form of an Ellipse

Foci of an Ellipse