For what values of k will the point (2,1) lies in the curve kx^2-2x^2+3y-3=0?
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Solution:
Given (2, -3) lies on kx2 – 3y2 + 2x + y – 2 = 0
Substitute (2, -3) in the given equation.
k×22 -3×32 + 2×2 – 3 – 2 = 0
=> 4k – 27 + 4 – 5 = 0
=> 4k = 28
=> k = 7
Hence option (3) is the answer.
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1
Step-by-step explanation:
answer is 7
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