TS ICET 23 May 2018 Shift 1 Paper

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Question : 78 of 200

Marks: +1, -0

4x^2 + 13y^2 + 9z^2 = 12y(x + z)

⇒ x : y : z =

1:2:3
2:3:1
4:6:9
9:6:4

Solution:

Given

4 x^{2}+13 y^{2}+9 z^{2}=12 y(x+z)

⇒ 4 x^{2}-12 x y+9 y^{2}+4 y^{2}-12y z+9 z^{2}=0

⇒(2 x-3 y)^{2}+(2 y-3 z)^{2}=0

Hence

2 x-3 y=0

and

2 y-3 z=0

⇒ x: y=y: z=3: 2

Required answer

x: y: z=9: 6: 4

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