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Question

Show that the equation (10x5)2+(10y5)2=(3x+4y1)2 represents an ellipse. Find the length of its latus rectum.

A
52
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B
12
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C
32
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D
12
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Solution

The correct option is C 32
(10x5)2+(10y5)2=(3x+4y1)2
25(2x1)2+25(2y1)2=(3x+4y1)2
(2x1)2+(2y1)2=(3x+4y15)2
4(x12)2+4(y12)2=14(3x+4y15)2
(x12)2+(y12)2=14(3x+4y15)2
By observing equation (1), we infer that
(x12)2+(y12)2=12(3x+4y15)
Centre of ellipse: (12,12)
(e2=1b2a2)
Equation of directrix =(3x+4y1=0)
Also d (Centre, directrix) =3(12)+4(12)15
=52+15=12
a+ae=12;a+a2=12
3a2=12;a13
1b2a2=14;b2a2=34;b=112
Length =2b2a=2(12)×11/3=12





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