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for what value of non negative integer a will the e system x²-y²=0,(x-a)²+y²=1 have exactly three real solutions

for what value of non negative integer a will the e system x²-y²=0,(x-a)²+y²=1 have exactly three real solutions

Grade:12

1 Answers

Arun
25750 Points
3 years ago
given =  x²-y²=0  
             (x-k)²+y²=1  => x²- 2kx + k²+y² =1
 
by adding this two equations we will get,
2x² - 2kx + k²= 0
 
and now we all know that if the discriminant is zero i.e., b² + 4ac=0 then the quadratic equation has one solution.
 
Then, 
           4k² - 8k = 0
           4k²=8k
           4k = 8
             k = 8/4 = 2 
 
so the correct answer is a) 2
 
 

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