Conic-Sections Question 566
Question: The equation of the ellipse with its centre at $ (1,2), $ focus at (6,2) and passing through the point $ (4,6) $ is $ \frac{{{(x-1)}^{2}}}{a^{2}}+\frac{{{(y-2)}^{2}}}{b^{2}}=1 $ , then
Options:
A) $ a^{2}=1,b^{2}=25 $
B) $ a^{2}=25,b^{2}=20 $
C) $ a^{2}=20,b^{2}=25 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Centre is (1, 2) and focus is (6, 2), hence the line joining the centre C and the focus S (i.e., the axis) is parallel to x-axis. Therefore, the equation must be of the form $ \frac{{{(x-1)}^{2}}}{a^{2}}+\frac{{{(y-2)}^{2}}}{b^{2}}=1 $ ?. (i) The distance between the centre and the focus $ CS=ae=5 $ Also $ b^{2}=a^{2}(1-e^{2})=a^{2}-a^{2}e^{2}=a^{2}-a^{2}e^{2} $
$ \therefore $ The equation (i) becomes $ \frac{{{(x-1)}^{2}}}{a^{2}}+\frac{{{(y-2)}^{2}}}{a^{2}-25}=1 $ ?. (ii) The point (4, 6) lies on it, therefore $ \frac{{{(4-1)}^{2}}}{a^{2}}+\frac{{{(6-2)}^{2}}}{a^{2}-25}=1\Rightarrow a^{2}=45 $
$ \therefore b^{2}=45-25=20. $ The required equation is $ \frac{{{(x-1)}^{2}}}{45}+\frac{{{(y-2)}^{2}}}{20}=1. $ $ 9ac-9a^{2}-2c^{2}>9ac-6c^{2} $ ? (i) Again $ 3a<2c\Rightarrow 9ac<6c^{2} $
$ \Rightarrow 9ac-6c^{2}<0 $ ? (ii) From (i) and (ii), $ 9ac-9a^{2}-2c^{2}>0 $
BETA
