Question

Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse $16x^2+y^2=17$

Answer 1

The given equation is $16x^2+y^2=16$
It can be written as
$16x^2+y^2=16$

or $\dfrac{x^2}{1} + \dfrac{y^2}{16} = 1$

or $\dfrac{x^2}{1^2} + \dfrac{y^2}{4^2} = 1$

Here, the denominator of $\dfrac{y^2}{4^2}$ is greater than the denominator of $\dfrac{x^2}{1^2}.$
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation (1) with $\dfrac{x^2}{b^2} \+ \dfrac{y^2}{a^2} = 1$, we obtain $b = 1$ and $a =4$.
$∴c = √(a^2 – b^2)$

$= √(16-1)$

$= √15$

Therefore,
The coordinates of the foci are $(0, ±√15).$
The coordinates of the vertices are $(0, ±4).$
Length of major axis = $2a= 8$
Length of the minor axis = $2b = 2$
Eccentricity,$e = \dfrac{c}{a} = \dfrac{√15}{4}$
Length of latus rectum = $\dfrac{2b^2}{a} = \dfrac{(2×12)}{4} = \dfrac{2}{4} = \dfrac{1}{2}$