Write an equation in standard form for each ellipse with center (0, 0) and co-vertex at (5, 0); focus at (0, 3).
Accepted Solution
A:
Answer:The required standard form of ellipse is [tex]\frac{x^2}{25}+\frac{y^2}{34}=1[/tex].Step-by-step explanation:The standard form of an ellipse is[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]Where, (h,k) is center of the ellipse.It is given that the center of the circle is (0,0), so the standard form of the ellipse is[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex] .... (1)If a>b, then coordinates of vertices are (±a,0), coordinates of co-vertices are (0,±b) and focus (±c,0).[tex]c^2=a^2-b^2[/tex] .... (2)If a<b, then coordinates of vertices are (0,±b), coordinates of co-vertices are (±a,0) and focus (0,±c).[tex]c^2=b^2-a^2[/tex] .... (3)It is given that co-vertex of the ellipse at (5, 0); focus at (0, 3). So, a<b we get[tex]a=5,c=3[/tex]Substitute a=5 and c=3 these values in equation (3).[tex]3^2=b^2-(5)^2[/tex][tex]9=b^2-25[/tex][tex]34=b^2[/tex][tex]\sqrt{34}=b[/tex]Substitute a=5 and [tex]b=\sqrt{34}[/tex] in equation (1), to find the required equation.[tex]\frac{x^2}{5^2}+\frac{y^2}{(\sqrt{34})^2}=1[/tex][tex]\frac{x^2}{25}+\frac{y^2}{34}=1[/tex]Therefore the required standard form of ellipse is [tex]\frac{x^2}{25}+\frac{y^2}{34}=1[/tex].