If P and Q are two different points on the cu | vedclass

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(D) Let $P = (t_1, t_1^3 - t_1)$ and $Q = (t_2, t_2^3 - t_2)$.The slope of the tangent at $P$ is given by $\frac{dy}{dx} = 3x^2 - 1$,so at $P$,$m = 3t

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$4$

Vedclass · Answer

(D) Let $P = (t_1, t_1^3 - t_1)$ and $Q = (t_2, t_2^3 - t_2)$.The slope of the tangent at $P$ is given by $\frac{dy}{dx} = 3x^2 - 1$,so at $P$,$m = 3t_1^2 - 1$.The slope of the chord $PQ$ is $\frac{(t_2^3 - t_2) - (t_1^3 - t_1)}{t_2 - t_1} = \frac{(t_2 - t_1)(t_2^2 + t_1t_2 + t_1^2) - (t_2 - t_1)}{t_2 - t_1} = t_2^2 + t_1t_2 + t_1^2 - 1$.Since the tangent at $P$ intersects the curve at $Q$,the slope of the tangent equals the slope of the chord $PQ$:$3t_1^2 - 1 = t_2^2 + t_1t_2 + t_1^2 - 1$$2t_1^2 - t_1t_2 - t_2^2 = 0$$(2t_1 + t_2)(t_1 - t_2) = 0$.Since $P$ and $Q$ are distinct,$t_1 
eq t_2$,so $t_2 = -2t_1$.The slope $m_{OP} = \frac{t_1^3 - t_1}{t_1} = t_1^2 - 1$ and $m_{OQ} = \frac{t_2^3 - t_2}{t_2} = t_2^2 - 1$.We need to calculate $\frac{m_{OQ} + 1}{m_{OP} + 1} = \frac{(t_2^2 - 1) + 1}{(t_1^2 - 1) + 1} = \frac{t_2^2}{t_1^2}$.Substituting $t_2 = -2t_1$,we get $\frac{(-2t_1)^2}{t_1^2} = \frac{4t_1^2}{t_1^2} = 4$.

If $P$ and $Q$ are two different points on the curve $y = x^3 - x$ such that the tangent at $P$ intersects the curve again at $Q$,then $\frac{m_{OQ} + 1}{m_{OP} + 1}$ is equal to,where $O$ is the origin and $m_{AB}$ represents the slope of the line segment $AB$.

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