EllipseHardBloom L3
Question
For the ellipse $5x^{2} + 9y^{2} + 10x - 36y - 4 = 0$ which of the following are true :
Options
A.centre is $(1,2)$
B.one of the foci is $( - 3,2)$
C.length of latus rectum is $\frac{10}{3}$
D.eccentricity is $\frac{2}{3}$
Solution
Sol. $\frac{(x + 1)^{2}}{9} + \frac{(y - 2)^{2}}{5} = 1$
centre $\equiv ( - 1,2)$
$$5 = 9\left( 1 - e^{2} \right) \Rightarrow e = \frac{2}{3}$$
Foci $\equiv \left( - 1 + 3\left( \frac{2}{3} \right),2 + 0 \right)$
and $\left( - 1 - 3\left( \frac{2}{3} \right),2 + 0 \right)$
Foci $\equiv (1,2)$ and $( - 3,2)$
Length of L.R. $= \frac{2b^{2}}{a} = 2\left( \frac{5}{3} \right)$
⇒ Length of L.R. $= \frac{10}{3}$
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