If two different tangents of y2=4ax are normals to x2=4y, then
A
|a|>2√2
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B
|a|<2√2
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C
|a|>12√2
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D
|a|<12√2
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Solution
The correct option is A|a|>2√2 The tangent to the parabola y2=4ax is y=m1x+am1...(1) The normal to the parabola x2=4y is x=m2y−2m2−(m2)3...(2) comparing equation (1) and (2) m1=1m2 & am1=2+(m2)2 ⇒(m2)2−am2+2=0 For two distinct tangents/normals D>0 a2−8>0 ⇒|a|>2√2