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Find the vertices and foci of the ellipse. 5x^(2)+9y^(2)=45

User Martijnve
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Sure, let's solve this step by step.

First, we must rewrite the equation in the standard form of an ellipse, which is x^(2)/a^(2) + y^(2)/b^(2) = 1. To do this, we'll divide each term by 45 to make the right side of the equation equal to 1:

5x^(2)/45 + 9y^(2)/45 = 1

This simplifies to:

x^(2)/3^(2) + y^(2)/5^(2) = 1

We interpret the constants 3 and 5 as the distances from the center of the ellipse to the vertices along the x and y axes, respectively. In an equation of an ellipse where the larger denominator under y, the major axis is vertical. Thus, the semi-major axis (a) is 5 and the semi-minor axis (b) is 3.

Therefore, the coordinates of the vertices are the points where the ellipse intersects the axes. These are at (0, ±a) and (±b, 0), which would mean our vertices are at (0, ±5) and (±3, 0). Upon solving, our vertices are (0, 5), (0, -5), (3, 0), and (-3, 0).

Next, let's calculate the foci. The distance from the center of the ellipse to each focus (c) is computed as the square root of the difference of the squares of a and b, or √(a^2 - b^2).

Let's calculate c using a = 5 and b = 3:

c = √(5^2 - 3^2)
c = √(25 - 9)
c = √16
c = 4

So, the coordinates of the foci are also points on the y-axis given by (0, ±c), resulting in our foci being at (0, +4) and (0, -4).

In conclusion, the vertices of the given ellipse are located at (0, 5), (0, -5), (3, 0), and (-3, 0). The foci are at (0, 4) and (0, -4).

User Gisgyaan
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