(x+1)^2/4 + (y-1)^2/9 = 11
Sketch the graph and find the vertices, asymptotes and foci.
I think that the center is (-1,1) and that y is the major axis.
Thanks so much!
In all probability, that is the case.
We have,
$\displaystyle \frac{(x + 1)^2}{4} + \frac{(y - 1)^2}{9} = 1$ ---- (i)
Since the ellipse is centred at $\displaystyle (-1,1)$, shifting the origin at $\displaystyle (-1,1)$ without rotating the coordinate axes, we have
$\displaystyle x = X - 1$ and $\displaystyle y = Y + 1$ ---- (ii)
Using these relations in (i), it reduces to
$\displaystyle \frac{X^2}{4} + \frac{Y^2}{9} = 1$ ---- (iii)
Clearly, this equation is of the form $\displaystyle \frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$, where $\displaystyle a^2 = 4$ and $\displaystyle b^2 = 9$ i.e. $\displaystyle a = 2$ and $\displaystyle b = 3$
We find that b > a. So the major and minor axes of the ellipse (iii) are along Y and X axes respectively.
$\displaystyle \therefore$ Length of the major axis = 2a = 4 ; Length of the minor axis = 2b = 6
The eccentricity $\displaystyle e$ is given by $\displaystyle e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$ ---- ●
Vertices:
The coordinates of the vertices with respect to the new axes are given by $\displaystyle (X = 0,\ Y = \pm b)$ i.e. $\displaystyle (X = 0,\ Y = \pm 3)$.
So, the vertices with respect to the old axes are given by
$\displaystyle (-1, 1 \pm 3)$ i.e. $\displaystyle \boxed{(-1,-2)}$ and $\displaystyle \boxed{(-1, 4)}$ [Putting $\displaystyle X = 0$, $\displaystyle Y = \pm 3$ in (ii)]
Foci:
The coordinates of the foci with respect to the new axes are given by
$\displaystyle (X = 0,\ Y = \pm be)$ i.e. $\displaystyle (X = 0,\ Y = \pm \sqrt{5})$ (Using ●)
So the coordinates of the foci with respect to the old axes are:
$\displaystyle (-1, 1\pm \sqrt{5})$ i.e $\displaystyle \boxed{(-1, 1 - \sqrt{5})}$ and $\displaystyle \boxed{(-1, 1 + \sqrt{5})}$
Asymptotes:
Before we find the equation of the asymptotes, recall the condition for a general equation of second degree $\displaystyle ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent a pair of straight lines -
$\displaystyle \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ ---- (1)
(NOTE: In the general equation, the coefficients of the xy, x and y terms are written as 2h, 2g and 2f only for the sake of convenience in calculation of $\displaystyle \Delta$.You can very well write the general equation as $\displaystyle ax^2 + bxy + cy^2 + dx + ey + f = 0$. In that case $\displaystyle \Delta = acf + \frac{edb}{4} - a\frac{e^2}{4} - b\frac{d^2}{4} - f\frac{b^2}{4} = 0$)
Coming back to our problem, since the equation to the asymptotes differs from the equation to the curve only in its constant term, the required equation of the asymptotes must be:
$\displaystyle ax^2 + 2hxy + by^2 + 2gx + 2fy + c + \lambda = 0$ ---- (2)
Also (2) is to be a pair of straight lines.
Hence, using (1),
$\displaystyle ab(c + \lambda) + 2fgh - af^2 - bg^2 - (c + \lambda)h^2 = 0$
Therefore $\displaystyle \lambda = - \frac{abc + 2fgh - af^2 - bg^2 - ch^2}{ab - h^2} = - \frac{\Delta}{ab - h^2}$
Thus the required equation of the asymptotes is:
$\displaystyle ax^2 + 2hxy + by^2 + 2gx + 2fy + c - \frac{\Delta}{ab - h^2} = 0$
In our case, the equation of the ellipse is
$\displaystyle \frac{(x + 1)^2}{4} + \frac{(y - 1)^2}{9} = 1$
$\displaystyle \Rightarrow 9(x + 1)^2 + 4(y - 1)^2 = 36$
$\displaystyle \Rightarrow 9x^2 + 4y^2 + 18x - 8y - 23 = 0$
Thus, we have a = 9, h = 0, b = 4, g = 9, f = - 4, c = - 23.
Using (1), $\displaystyle \Delta = \left(\frac{9\times 4\times -23 + 0 - 9\times 16 - 4\times 81}{9\times 4}\right) = - 1296$
So the required equation of the asymptotes is:
$\displaystyle 9x^2 + 4y^2 + 18x - 8y - 23 - \frac{-1296}{36} = 0$
$\displaystyle \Rightarrow \boxed{9x^2 + 4y^2 + 18x - 8y + 13 = 0}$
NOTE: You must know that the equation of asymptotes of an ellipse give an imaginary pair of straight lines. You may write the equation in terms of y, which gives $\displaystyle y = \frac{8 \pm \sqrt{-(144x^2 + 288x + 144)}}{8}$. The term under the root is always negative. Further, you can try to graph the asymptotes using graphing software. They won't appear!