If three distinct and real normals can be drawn to y2=8x from the point (a,0), then-
A
a>2
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B
a∈(2,4)
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C
a>4
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D
none of these
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Solution
The correct option is Da>4 Equation of normal to the parabola is given by, y+tx=2at+at3 Here a=2 so, y+tx=4t+2t3 Given it passes through (a,0) ⇒ta=4t+2t3⇒t(2t2−a+4)..(1) Therefore for three distinct normal through (a,0) equation (1) must have three distinct roots ⇒−a+4<0⇒a>4