Consider P is a point on y2=4ax, if the normal at P, the axis and the focal radius of P form an equilateral triangle. Then coordinates of P are
A
(4a,4a)
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B
(3a,2√3a)
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C
(a3,2a√3)
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D
(3a,−2√3a)
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Solution
The correct option is C(a3,2a√3) Let P be (at2,2at) and PN be normal √32SP=PM, where M is foot of perpendicular from P upon axis of parabola ⇒√32(at2+a)=2at ⇒√3t2+√3=4t ⇒√3t2−4t+√3=0 ⇒√3t2−3t−t+√3 ⇒√3t(t−√3)−1(t−√3)=0 ⇒(t−√3)(√3t−1)=0 ∴t=√3 or t=1√3
So, P is (3a,2√3a)or(a3,2a√3)