If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

y = x3 d y d x = 3 x 2 ⇒ d y d x | ( t , t 3 ) = 3 t 2 Equation of tangent y – t3 = 3t2 (x – t) Let again meet the curve at Q ( t 1 , t 1 3 )  ⇒ t 1 3 − t 3 = 3 t 2 ( t 1 − t )  t 1 2 + t t 1 + t 2 = 3 t 2 [ ? t 1 ≠ t ]  t 1 2 + t t 1 − 2 t 2 = 0 => t1 = -2tRequired ordinate = 2 t 3 + t 1 3 3 = 2 t 3 − 8 t 3 3 = − 2 t 3

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Applications of Derivatives Maths Applications of Derivatives Maths Class 12th

If the tangent to the curve y = x³ at the point P(t, t³) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

Option 1 -

Option 2 -

-t³

Option 3 -

-2t³

Option 4 -

2t³

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 weeks ago

Correct Option - 3

Detailed Solution:

y = x³

${\frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} |}_{(t, t^{3})} = 3 t^{2}$

Equation of tangent y – t³ = 3t² (x – t)

Let again meet the curve at $Q (t_{1}, t_{1}^{3})$

$\Rightarrow t_{1}^{3} - t^{3} = 3 t^{2} (t_{1} - t)$

$t_{1}^{2} + t t_{1} + t^{2} = 3 t^{2} [? t_{1} \neq t]$

$t_{1}^{2} + t t_{1} - 2 t^{2} = 0$

=> t₁ = -2t

Required ordinate = $\frac{2 t^{3} + t_{1}^{3}}{3} = \frac{2 t^{3} - 8 t^{3}}{3} = - 2 t^{3}$

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If the tangent to the curve y = x³ at the point P(t, t³) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

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