Question

Find the ends of the major axisand foci.49x2 + 16y2 = 784

Answer 1

The general equation of an ellipse is:

`\frac{\mleft(x-h\mright)^2^{}}{b^2}+((y-k)^2)/(a^2)=1`

where a > b.

To get this form, we need to divide by 49 and by 16, the equation provided, as follows:

`\begin{gathered} 49x^2+16y^2=784 \\ (49x^2+16y^2)/(49\cdot16)=(784)/(49\cdot16) \\ (49x^2)/(49\cdot16)+(16y^2)/(49\cdot16)=(784)/(784) \\ (x^2)/(16)+(y^2)/(49)=1 \end{gathered}`

This means that the values of the constants are: h = 0, k = 0, b = 4, and a = 7.

Since a is greater than b, then the major axis is the vertical axis. The ends of the major axis are the vertices and they are found as follows:

(h, k+a) = (0, 0+7) = (0, 7)

(h, k-a) = (0, 0-7) = (0, -7)

And the foci are found as follows:

(h, k+c) = (0, 0+5.74) = (0, 5.74)

(h, k-c) = (0, 0-5.74) = (0, -5.74)

where c is computed as follows:

`\begin{gathered} c^2=a^2-b^2 \\ c^2=49-16 \\ c=\sqrt[]{33} \\ c\approx5.74 \end{gathered}`