Question:
If the tangent at a point $P$, with parameter $t$, on the curve $x = 4t^2 + 3, y = 8t^3 - 1, t \in R$, meets the curve again at a point $Q$, then the coordinates of $Q$ are :
If the tangent at a point $P$, with parameter $t$, on the curve $x = 4t^2 + 3, y = 8t^3 - 1, t \in R$, meets the curve again at a point $Q$, then the coordinates of $Q$ are :
Updated On: Apr 24, 2026
- $(t^2 +3 , -t^3 -1)$
- $(4t^2 + 3, -8t^3 -1)$
- $(t^2 + 3, t^3 -1)$
- $(16t^2 +3 ,-64 t^3 - 1)$
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The Correct Option is A
Solution and Explanation
$P\left(^{A}t^{2}, \,3\, 8t^{3}\, 1\right)$
$\frac{du/dt}{dx/dt}=\frac{dy}{dx}=3t$ (slope of tangent at P)
Let $Q=\left(4\lambda^{2}+3,8\lambda^{3}-1\right)$
slope of $PQ = 3t$
$\frac{8t^{3}-8\lambda ^{3}}{4t^{2}-4\lambda ^{2}}=3t$
$\Rightarrow t^{2}+t\lambda-2\lambda^{2}=0$
$\left(t-\lambda\right)\left(t+2\lambda\right)=0$
$t=\lambda\left(or\right)\lambda=\frac{-t}{2}$
$\therefore Q=\left[t^{2}+3,t^{3}-1\right]$
$\frac{du/dt}{dx/dt}=\frac{dy}{dx}=3t$ (slope of tangent at P)
Let $Q=\left(4\lambda^{2}+3,8\lambda^{3}-1\right)$
slope of $PQ = 3t$
$\frac{8t^{3}-8\lambda ^{3}}{4t^{2}-4\lambda ^{2}}=3t$
$\Rightarrow t^{2}+t\lambda-2\lambda^{2}=0$
$\left(t-\lambda\right)\left(t+2\lambda\right)=0$
$t=\lambda\left(or\right)\lambda=\frac{-t}{2}$
$\therefore Q=\left[t^{2}+3,t^{3}-1\right]$
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Application of Derivatives
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