Question

consider the following equation of an ellipse.

25x² + 49y² - 200x + 294y - 384 = 0

step 3 of 4: find the endpoints of the major and minor axes of this ellipse.

Explanation:

Step1: Rewrite the equation

Group \(x\) - terms and \(y\) - terms:

$$
LATEXBLOCK0
$$

Complete the square for \(x\) and \(y\):
For \(x\): \((x - 4)^{2}=x^{2}-8x + 16\), so \(25(x^{2}-8x)=25[(x - 4)^{2}-16]\)
For \(y\): \((y + 3)^{2}=y^{2}+6y+9\), so \(49(y^{2}+6y)=49[(y + 3)^{2}-9]\)
The equation becomes:

$$
LATEXBLOCK1
$$

The standard form of an ellipse is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) (\(a>b>0\)), here \(h = 4,k=-3,a = 7,b = 5\)

Step2: Find endpoints of major and minor axes

For the major axis (parallel to the \(x\) - axis since \(a^{2}=49\) is under the \(x\) - term):
The endpoints are \((h\pm a,k)\)
Substitute \(h = 4,k=-3,a = 7\)
\((4 + 7,-3)=(11,-3)\) and \((4-7,-3)=(-3,-3)\)

For the minor axis (parallel to the \(y\) - axis):
The endpoints are \((h,k\pm b)\)
Substitute \(h = 4,k=-3,b = 5\)
\((4,-3 + 5)=(4,2)\) and \((4,-3-5)=(4,-8)\)

Answer:

Endpoints of major axis: \((11,-3)\) and \((-3,-3)\)
Endpoints of minor axis: \((4,2)\) and \((4,-8)\)