Let x=my+c is normal to x2=4y, if k2+mk+m=0 has only one real value of k,then value(s) of c is/are
A
0
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B
72
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C
−72
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D
64
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Solution
The correct option is C−72 Using the slope form of the normal for the given parabola, we get x=my−2m−m3 ∴c=−2m−m3
Now k2+mk+m=0 will have equal roots if m2−4m=0⇒m=0,4
Hence values of c=0,−72