Find the Vertex Form (x^2)/4+(y^2)/25=1

Problem

(x^2)/4+(y^2)/25=1

Solution

1. Identify the type of conic section. The equation is in the standard form of an ellipse centered at the origin (0,0) because it follows the form (x^2)/(a^2)+(y^2)/(b^2)=1

2. Determine the values of a^2 and b^2 From the equation, we see that the denominator under x^2 is 4 and the denominator under y^2 is 25

3. Calculate the semi-axes lengths. Since b^2=25 is greater than a^2=4 the ellipse is vertically oriented.

a=√(,4)=2

b=√(,25)=5

4. Locate the vertices. For a vertical ellipse centered at (h,k) the vertices are located at (h,k±b) Here, (h,k)=(0,0) and b=5

(V_1)=(0,5)

(V_2)=(0,−5)

5. Locate the co-vertices. The co-vertices are located at (h±a,k) Here, a=2

(C_1)=(2,0)

(C_2)=(−2,0)

6. Express in vertex form. For an ellipse, the "vertex form" is the standard form showing the center (h,k)

((x−h)^2)/(a^2)+((y−k)^2)/(b^2)=1

Final Answer

(x^2)/4+(y^2)/25=1⇒((x−0)^2)/2^2+((y−0)^2)/5^2=1

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