A normal is drawn at the point P to the parabola y 2 = 8 x , which is inclined at 6 0 ∘ with the straight line y = 8 . Then the point P lies on the straight line

For the parabola y 2 = 4 a x , the equation of normal at P ( a m 2 , − 2 am ) is y = m x − 2 am − a m 3 . Here, m = tan 6 0 ∘ = 3 ∴ P ≡ ( a ( 3 ) 2 , − 2 a ( 3 ) ) ≡ ( 6 , − 4 3 ) ( ∵ a = 2 ) Thus, P satisfies 2 x − y − 12 − 4 3 = 0

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MathematicsParabolaMiscellaneous/MixedMedium2 minPYQ_2022

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A normal is drawn at the point $P$ to the parabola $y^{2} = 8 x$ , which is inclined at $6 0^{\circ}$ with the straight line $y = 8$ . Then the point $P$ lies on the straight line

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For the parabola $y^{2} = 4 a x$ , the equation of normal at $P (a m^{2}, - 2 am)$ is $y = m x - 2 am - a m^{3}$ . Here, $m = tan 6 0^{\circ} = 3$ $∴ P \equiv (a (3)^{2}, - 2 a (3)) \equiv (6, - 43) (∵ a = 2)$ Thus, $P$ satisfies $2 x - y - 12 - 43 = 0$

Stream:BITSATSubject:MathematicsTopic:ParabolaSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2022

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