If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3-1, t [ ] …

Question

If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3-1, t $$ \in $$ R, meets the curve again at a point Q, then the coordinates of Q are :

Accepted Answer

Given, x = 4t2 + 3 and y = 8t3 - 1

$$ \therefore $$   P $$ \equiv $$ (4t2 + 3, 8t3 - 1)

$${{dx} \over {dt}} = 8t$$ and $${{dy} \over {dt}}$$ $=$ 24t2

Slope of tangent at

P $=$ $${{dy} \over {dx}}$$ $=$ $${{dy/dt} \over {dx/dt}}$$ $=$ 3t

Assume Q $=$ (4$$\lambda $$2 + 3, 8$$\lambda $$3 - 1)

$$ \therefore $$   Slope of PQ $=$ 3t

$$ \Rightarrow $$     $${{8{\lambda ^3} - 8{t^3}} \over {4{\lambda ^2} - 4{t^2}}} = 3t$$

$$ \Rightarrow $$   $${{8\left( {\lambda  - t} \right)\left( {{\lambda ^2} + {t^2} + \lambda t} \right)} \over {4\left( {\lambda  - t} \right)\left( {\lambda  + t} \right)}} = 3t$$

$$ \Rightarrow $$    2$$\lambda $$2 + 2t2 + 2$$\lambda $$t $=$ 3t (t + $$\lambda $$)

$$ \Rightarrow $$    t2 + $$\lambda $$t - 2$$\lambda $$2 = 0

$$ \Rightarrow $$   (t - $$\lambda $$) (t + 2$$\lambda $$) $=$ 0

$$ \therefore $$   $$\lambda $$ $=$ t   or   $$\lambda $$ $=$ - $${t \over 2}$$

$$ \therefore $$   Q $=$  (4t2 + 3,  8t3 - 1)  Or

Q $=$ $$\left( {4 \times {{\left( { - {t \over 2}} \right)}^2} + 3, - 8 \times {{{t^3}} \over 8} - 1} \right)$$

$=$ (t2 + 3, - t3 - 1)

Solutions · 1