A point moves so that the sum of it's distances from (ae, 0) and (-ae, 0) is 2a. Prove that equation of it's
locus be
+
= 1, where b? = a'(1-e)
Answers
Answer:
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MATHS
A points moves that the sum of its distances from two fixed points (ae , 0) and (-ae , 0 ) is always 2a . Prove that the equation of the locus is
a
2
x
2
+
a
2
(1−e
2
)
y
2
=1
ANSWER
Suppose that the co-ordinates of the point are (x , y), then the given condition
[(x−ae)
2
+y
2
]
+
[(x+ae)
2
+y
2
]
=2a .........(1)
Now
[(x−ae)
2
+y
2
]−[(x+ae)
2
+y
2
]=−4aex ............(2)
[∵(a−b)
2
−(a+b)
2
=−4ab]
On dividing (2) by (1), we get
[(x−ae)
2
+y
2
]
−
[(x+ae)
2
+y
2
]
=−2ex
[∵
(
L
)+(
M
)
L−M
=(
L
)−(
M
)]
Adding (1) and (3) we get
2
[(x−ae)
2
+y
2
]
=2(a−ex). Square
x
2
−2aex+a
2
e
2
+y
2
=a
2
−2aex+e
2
x
2
or x
2
(1−e
2
)+y
2
=a
2
(1−e
2
)
or
a
2
x
2
+
a
2
(1−e
2
)
y
2
=1
Note : The above method will be referred to as L-M method throughout this book.
Answered By
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I.I.T Maths M.L Khanna Chapter 17 Set 1
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