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Question

If the focus of the parabola x2−ky+3=0 is (0,2) then the values of k are?

A
32
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B
3
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C
4
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D
6
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Solution

The correct option is A 32
focusofparabolax2ky+3=0is(0,2),thenvalueofkbyputtingx=0andy=2ineqn0k×2+3=02k+3=02k=3,k=32

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