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NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Parabola

Let the tangents at points& P and R on the parabola y = x^ 2...

Let the tangents at points&

P

and

R

on the parabola

y = x^{2}

intersects at

T .

Tangent at point

Q

(lies in between the points

P

and

R

) on the parabola intersect

PT

and

RT

at

A

and

B

respectively. The value of

\dfrac{TA}{TP} + \dfrac{TB}{TR}

& is

Options

A $\dfrac{1}{2}$
B $1$
C $2$
D $4$

Correct answer

B. $1$

Step-by-step solution

Let,& $P =$ \left({t}_{1}, {t}_{1}^{2}\right), $Q = \left({t}_{2}, {t}_{2}^{2}\right) & R = \left({t}_{3}, {t}_{3}^{2}\right)<br />$ ⇒ $\dfrac{TA}{TP} = \left(\dfrac{{t}_{1} + {t}_{3}}{2}\right) - \left(\dfrac{{t}_{1} + {t}_{2}}{2}\right)\left(\dfrac{{t}_{1} + {t}_{3}}{2} - {t}_{1}\right) = \dfrac{{t}_{3} - {t}_{2}}{{t}_{3} - {t}_{1}} and</p><p>\dfrac{TB}{TR} = \dfrac{{t}_{1} - {t}_{2}}{{t}_{1} - {t}_{3}} = \dfrac{{t}_{2} - {t}_{1}}{{t}_{3} - {t}_{1}}<br />$ ⇒ $\dfrac{TA}{TP} + \dfrac{TB}{TR} = \dfrac{{t}_{3} - {t}_{2}}{{t}_{3} - {t}_{1}} + \dfrac{{t}_{2} - {t}_{1}}{{t}_{3} - {t}_{1}} = \dfrac{{t}_{3} - {t}_{1}}{{t}_{3} - {t}_{1}} =1$

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