Question

Previous year Question of jee mains
Chapter:- application and derivatives​

Answer 1

EXPLANATION.

Tangent to the curve y = x³ at the point P(t, t³) meet the curve again at Q.

The ordinates of the point which divides PQ internally in the ratio 1 : 2.

As we know that,

⇒ P = (t, t³).

⇒ Q = (t₁, t₁³).

As we know that,

Slope of the line = (y₂ - y₁)/(x₂ - x₁).

Using this formula in the equation, we get.

Slope of the PQ = (t₁³ - t³)/(t₁ - t).

⇒ (t₁ - t)(t₁² + t₁t + t²)/(t₁ - t) = (t₁² + t₁t + t²).

Equation of the curve : y = x³.

dy/dx at point (t, t³) = 3t².

dy/dx = Slope of the tangent.

⇒ t₁² + t₁t + t² = 3t².

⇒ t₁² + t₁t + t² - 3t² = 0.

⇒ t₁² + t₍t - 2t² = 0.

⇒ (t₁ + 2t)(t₁ - t) = 0.

⇒ t₁ ≠ t.

⇒ t₁ = - 2t.

Co-ordinates of Q = (t₁, t₁³).

Co-ordinates of Q = (-2t, -8t³).

Ordinates of the point = (2t³ - 8t³)/2 + 1 = -6t³/3 = -2t³

Option [1] is correct answer.

Answer 2

$\Large{\underline{\underline{\bold{☆ Given:}}}}$

• Tangent in the curve y = x³ at the point P(t,t³) meets the curve again at point Q.

$\Large{\underline{\underline{\red{\bold{➸To\;Find:}}}}}$

• Ordinate of the point that divides PQ in the ratio of 1:2.

$\Large{\underline{\underline{\bold{๛ Required\; Response:}}}}$

Slope of the tangent at point p

$\sf{P(t,t³) = \frac{dy}{dx}]_{(t,t³)}}$

= (3x²)x = t = 3t²

So equation tangent at P(t,t³)

$y - {t}^{3} = 3 {t}^{2} (x - t)$

For point of intersection with y = x³

${x}^{3} - {t}^{3} = 3 {t}^{2} x - 3 {t}^{3} \\ (x - t)( {x}^{2} + xt + {t}^{2} ) = 3 {t}^{2} (x - t)$

For x ≠ t

${x}^{2} + xt + {t}^{2} = 3 {t}^{2} \\ {x}^{2} + xt + {t}^{2} - 3 {t}^{2} = 0 \\ {x}^{2} + xt - 2 {t}^{2} = 0 \\ (x - t)(x + 2t) = 0$

For Q : x = -2t, Q(-2t,-8t³)

Required Point

$→\;\frac{2 {t}^{3} - 8 {t}^{3} }{2 + 1} \\ →\; \frac{ - 6 {t}^{3} }{3} \\→\; - 2 {t}^{3}$

Ordinate of the point which the line PQ in the ratio of 1:2 ☞ $\Large\boxed{\blue{-2t³}}$

• Option A.

$\boxed{\tt{@MrSovereign}}$

Hope This Helps!!