Find the eccentricity, center, foci and vertices of the ellipse $\;$ $x^2 + 4y^2 -8x - 16 y - 68 = 0$ $\;$ and sketch it.
Equation of the given ellipse is: $\;$ $x^2 + 4y^2 - 8x - 16 y - 68 = 0$
i.e. $\;$ $\left(x^2 - 8x\right) + 4 \left(y^2 - 4y\right) = 68$
i.e. $\;$ $\left[\left(x^2 - 8x + 16\right) - 16\right] + 4 \left[\left(y^2 - 4y + 4\right) - 4\right] = 68$
i.e. $\;$ $\left(x - 4\right)^2 + 4 \left(y - 2\right)^2 = 68 + 16 + 16$
i.e. $\;$ $\left(x - 4\right)^2 + 4 \left(y - 2\right)^2 = 100$
i.e. $\;$ $\dfrac{\left(x - 4\right)^2}{100} + \dfrac{\left(y - 2\right)^2}{100 / 4} = 1$
i.e. $\;$ $\dfrac{\left(x - 4\right)^2}{100} + \dfrac{\left(y - 2\right)^2}{25} = 1$ $\;\;\; \cdots \; (1)$
Let $X = x - 4$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 2$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ can be written as: $\;$ $\dfrac{X^2}{100} + \dfrac{Y^2}{25} = 1$ $\;\;\; \cdots \; (3)$
Major axis of the ellipse given by equation $(3)$ is along the X axis.
Comparing equation $(3)$ with the standard equation of ellipse $\;$ $\dfrac{X^2}{a^2} + \dfrac{Y^2}{b^2} = 1$ $\;$ gives
$a^2 = 100$ $\implies$ $a = 10$; $\;\;$ $b^2 = 25$ $\implies$ $b = 5$
$\therefore$ $\;$ Eccentricity $= e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{25}{100}} = \sqrt{\dfrac{75}{100}} = \dfrac{\sqrt{3}}{2}$
Now, $a \times e = 10 \times \dfrac{\sqrt{3}}{2} = 5 \sqrt{3}$
| Referred to X, Y | Referred to x, y $X = x - 4$; $\;$ $Y = y - 2$ i.e. $\;$ $x = X + 4$; $\;$ $y = Y + 2$ |
|
|---|---|---|
| Center | $\left(0, 0\right)$ | $X = 0 \implies x = 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ Center $= \left(4, 2\right)$ |
| Foci | $F_1 = \left(ae, 0\right)$ $F_1 = \left(10 \times \dfrac{\sqrt{3}}{2}, 0\right)$ i.e. $\;$ $F_1 = \left(5 \sqrt{3}, 0\right)$ |
$X = 5 \sqrt{3} \implies x = 5 \sqrt{3} + 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $F'_1 = \left(4 + 5 \sqrt{3}, 2\right)$ |
| $F_2 = \left(-ae, 0\right)$ i.e. $\;$ $F_2 = \left(-10 \times \dfrac{\sqrt{3}}{2}, 0\right)$ i.e. $\;$ $F_2 = \left(-5 \sqrt{3}, 0\right)$ |
$X = -5 \sqrt{3} \implies x = -5 \sqrt{3} + 4$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $F'_2 = \left(4 - 5 \sqrt{3}, 2\right)$ |
|
| Vertices | $A_1 = \left(+a, 0\right)$ i.e. $\;$ $A_1 = \left(10, 0\right)$ |
$X = 10 \implies x = 14$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $A'_1 = \left(14, 2\right)$ |
| $A_2 = \left(-a, 0\right)$ i.e. $\;$ $A_2 = \left(-10, 0\right)$ |
$X = -10 \implies x = -6$ $Y = 0 \implies y = 2$ $\therefore$ $\;$ $A'_2 = \left(-6, 2\right)$ |
