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2x²/a² - y²/b² -z²/c² =0
2x²/a² = y²/b² + z²/c² ------(1)
-x²/a² -y²/b² + 2z²/c² =0
2z²/c² = y²/b² + x²/a² ------(2)
-x²/a² +2y²/b² -z²/c² =0
2y²/b² = x²/a² + z²/c² -----(3)
if we see deeply , then we find that
x²/a² , y²/b² , and z²/c² are in AP
y²/b², z²/c² and x²/a² are in AP
z²/c², x²/a², and y²/b² are in AP
so, we conclude that ,
x²/a² = y²/b² = z²/c² = K (let )
x² = a² K => x = ±a√k
y² = b²K => y= ± b√k
z² = c²K => z = ±c √k
so, solution will be ,
( +, + , +)
( + , + , -)
(+, -, -)
( -, -, -)
(-, -, +)
( -, +, +)
( +, -, +)
(-, + , -)
so, number of solutions = 8
2x²/a² = y²/b² + z²/c² ------(1)
-x²/a² -y²/b² + 2z²/c² =0
2z²/c² = y²/b² + x²/a² ------(2)
-x²/a² +2y²/b² -z²/c² =0
2y²/b² = x²/a² + z²/c² -----(3)
if we see deeply , then we find that
x²/a² , y²/b² , and z²/c² are in AP
y²/b², z²/c² and x²/a² are in AP
z²/c², x²/a², and y²/b² are in AP
so, we conclude that ,
x²/a² = y²/b² = z²/c² = K (let )
x² = a² K => x = ±a√k
y² = b²K => y= ± b√k
z² = c²K => z = ±c √k
so, solution will be ,
( +, + , +)
( + , + , -)
(+, -, -)
( -, -, -)
(-, -, +)
( -, +, +)
( +, -, +)
(-, + , -)
so, number of solutions = 8
abhi178:
i hope this will correct
plz , check another questions
Answered by
1
Answer:
2x²/a² - y²/b² -z²/c² =0
2x²/a² = y²/b² + z²/c² ------(1)
-x²/a² -y²/b² + 2z²/c² =0
2z²/c² = y²/b² + x²/a² ------(2)
-x²/a² +2y²/b² -z²/c² =0
2y²/b² = x²/a² + z²/c² -----(3)
if we see deeply , then we find that
x²/a² , y²/b² , and z²/c² are in AP
y²/b², z²/c² and x²/a² are in AP
z²/c², x²/a², and y²/b² are in AP
so, we conclude that ,
x²/a² = y²/b² = z²/c² = K (let )
x² = a² K => x = ±a√k
y² = b²K => y= ± b√k
z² = c²K => z = ±c √k
so, solution will be ,
( +, + , +)
( + , + , -)
(+, -, -)
( -, -, -)
(-, -, +)
( -, +, +)
( +, -, +)
(-, + , -)
so, number of solutions = 8
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