# Find the equation of the ellipse if the centre is$(3 , -1 ),$ one of the foci is $(6 , -1)$ and passing through the point $(8 , -1)$.

Toolbox:
• Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
• $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 • http://clay6.com/mpaimg/2_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/2_2_Toolbox1.png
• Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
• End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae) and (\pm \large\frac{b^2}{a}$$,-ae)$
• Directrices $y=\pm \large\frac{a}{e}$.
• The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
• General form of standard ellipses with centre $C$ with major axis $2a$,minor axis $2b$ and axes parallel to the coordinate axes.
• $\large\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}$$=1 • In this case major axis y=k and minor axis x=h. • \large\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}$$=1$
• In this case major axis $x=h$ and minor axis $y=k$.
Step 1:
$C(3,-1)$ is the centre,$F(6,-1)$ is one focus and the ellipse passes through $P(8,-1).$
$CF$ is the major axis (i.e)$y=-1$ (parallel to the $x$-axis)
Now the point $P$ lies on $y=-1$.
Therefore it is one of the vertices.
Step 2:
$CF=ae$
$LP=a$
Therefore $ae=3$
$a=5$
$e=\large\frac{3}{5}$
Step 3:
$b=5\sqrt{1-\large\frac{9}{25}}$
$\Rightarrow 5.\large\frac{4}{5}$$=4 Step 4: The equation of the ellipse is \large\frac{(x-3)^2}{25}+\large\frac{(y+1)^2}{16}$$=1$