Let normals to the parabola y ^ 2 = 4x at variable points P( t _ 1 ^ 2 , 2 t _ 1 ) and Q( t _ 2 ^ 2 , 2 t _ 2 ) meet at the point R ( t ^ 2 , 2t), then the line joining P and Q always passes through a fixed point (α, β)

Correct answer: 2 t = - t _ 1 - 2 t _ 1 and t = - t _ 2 - 2 t _ 2 ⇒ t _ 1 , t _ 2 are the roots of the equation t = - x - (2)/(x) ⇒ x ^ 2 + tx + 2 = 0 has roots t _ 1 and t _ 2 ⇒ t _ 1 + t _ 2 = - t and t _ 1 t _ 2 = 2 Now, the equation of PQ is ( t _ 1 + t _ 2 )y = 2x + 2 t _ 1 t _ 2 ⇒ - ty = 2x + 4 ⇒ 2(x + 2) + ty = 0 ⇒ PQ always passes through the common point of x + 2 = 0 and y = 0 ⇒ (α, β) = (- 2,0) ⇒ αβ = 0 & α + β = - 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Parabola

NTA Abhyas 2020 · Mathematics · Parabola

Let normals to the parabola y ^ 2 = 4x at variable points P( t _ 1...

Let normals to the parabola

{y}^{2} = 4x

at variable points

P\left({t}_{1}^{2}, 2{t}_{1}\right)

and Q\left({t}_{2}^{2}, 2{t}_{2}\right)

meet at the point

\left({t}^{2}, 2t\right),

then the line joining

P and Q

always passes through a fixed point \left(\alpha, \beta\right)

, then the value of \left|\alpha + \beta\right|

is equal to

Correct answer

Step-by-step solution

t = - {t}_{1} - \dfrac{2}{{t}_{1}} $and$ t = - {t}_{2} - \dfrac{2}{{t}_{2}} $ $ ⇒ {t}_{1}, {t}_{2} $are the roots of the equation$ t = - x - \dfrac{2}{x} $$ ⇒ {x}^{2} + tx + 2 = 0 $has roots$ {t}_{1} \text{ $and } {t}_{2} $ ⇒ {t}_{1} + {t}_{2} = - t and {t}_{1}{t}_{2} = 2 $ Now, the equation of$ PQ $is \left({t}_{1} + {t}_{2}\right)y = 2x + 2{t}_{1}{t}_{2}$
$⇒ - ty = 2x + 4 ⇒ 2\left(x + 2\right) + ty = 0$
$⇒ PQ$ always passes through the common point of $x + 2 = 0 and y = 0$
$⇒ \left(\alpha, \beta\right) = \left(- 2,0\right) ⇒ \alpha\beta = 0$ & $\alpha + \beta = - 2$

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