JEE Mains · Maths · STD 12 - 6. Application of derivatives
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
- A \((16t^2 +3, - 64t^3 - 1)\)
- B \((4t^2 + 3, - 8t^3 - 2)\)
- C \((t^2 + 3,\,t^3 - 1)\)
- D \((t^2 + 3, - t^3 - 1)\)
Answer & Solution
Correct Answer
(D) \((t^2 + 3, - t^3 - 1)\)
Step-by-step Solution
Detailed explanation
\(P\left( {4{t^2} + 3,8{t^3} - 1} \right)\) \(\frac{{dy/dt}}{{dt/dt}} = \frac{{dy}}{{dx}} = 3t\) (slope of tangent at \(P\)) Let \(Q = \left( {4{\lambda ^2} + 3,8{\lambda ^3} - 1} \right)\) slope of \(PQ = 3t\) \(\frac{{8{t^3} - 8{\lambda ^3}}}{{4{t^2} - 4{\lambda ^2}}} = 3t\)…
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