Q-55- If real vale of x and y stisfies the equation x2-4y2-8x-12-0- then-A- 0 lt- y lt- 1-B- 2 lt- y lt- 6-C- -1 -8804-y -8804-1-D- - 2 lt- y lt- 6

X2+4y2 8x+12=0x2 8x+4y2+12=0x2 2 #215;4 #215;x+16+4y2 4=0x 42+4y2 4=0x 42=41 y2L.H.S. #160;is #160;a #160;perfect #160;squareL.H.S.=x 42 #8805;0Th ...

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Byju's Answer

Standard XII

Mathematics

General Solution of tan theta = tan alpha

Q.55. If real...

Question

Q.55. If real vale of x and y stisfies the equation $x^{2} + 4 y^{2} - 8 x + 12 = 0$ . then-
(A) 0 < y < 1
(B) 2 < y < 6
(C) $- 1 \leq y \leq 1$
(D) - 2 < y < 6

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Solution

$x^{2} + 4 y^{2} - 8 x + 12 = 0 x^{2} - 8 x + 4 y^{2} + 12 = 0 x^{2} - 2 \times 4 \times x + 16 + 4 y^{2} - 4 = 0 {(x - 4)}^{2} + 4 y^{2} - 4 = 0 {(x - 4)}^{2} = 4 (1 - y^{2}) L . H . S . i s a p e r f e c t s q u a r e L . H . S . = {(x - 4)}^{2} \geq 0 T h e r e f o r e, f o r s o l u t i o n t o e x i s t 4 (1 - y^{2}) \geq 0 y^{2} - 1 \leq 0 y^{2} - 1^{2} \leq 0 (y - 1) (y + 1) \leq 0 S o l u t i o n o f e q u a t i o n (x - a) (x - b) \leq 0, w h e r e a < b, i s g i v e n b y x \in [a, b] y \in [- 1, 1] - 1 \leq y \leq 1$

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