Question
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1$

Answer

The given equation is $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1$ or $\frac{\text{x}^2}{4^2}+\frac{\text{y}^2}{3^2}=1$ Here, the denominator of $\frac{\text{x}^2}{16}$ is greater than the denominator of $\frac{\text{y}^2}{9}$. Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. On comparing the given equation with $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{a}^2}=1,$ we obtain b = 4 and a = 3. $\therefore \text{c}=\sqrt{\text{a}^2-\text{b}^2}=\sqrt{16-9}=\sqrt{7}$ The coordinates of the foci are $(\pm\sqrt{7},0)$ The coordinates of the vertices are $(\pm4,0)$ Length of major axis = 2a = 8 Length of minor axis = 2b = 6 Eccentricity, $\text{e}=\frac{\text{c}}{\text{a}}=\frac{\sqrt{7}}{4}$ Lenght of lacus rectum $=\frac{2\text{b}^2}{\text{a}}=\frac{2\times9}{4}=\frac{9}{2}.$

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