Question

If $y=3t-t^3,x=3t^2$ satisfies $A{y}_2(y-\sqrt{3x})=x+3$ then $A$ is equal to

Answer 1

$y=3t-t^3-----\left(1\right)and$

$x=3t^2------\left(2\right)differentiatingw.r.t.tweget,\Rightarrow \frac{dy}{dt}=3-3t^2;\frac{dx}{dt}=6t\Rightarrow \frac{dy}{dx}=\frac{3-3t^2}{6t}=\frac{1-t^2}{2t}(\frac{f}{g}{)}^{\prime }=\frac{f^{\prime }g-f{g}^{\prime }}{g^2}\frac{d^{2}y}{d{x}^2}=\frac{-2t\left(2t\right)-2(1-t^2)}{4t^2}\times \frac{dt}{dx}=\frac{-1-t^2}{2t^2}\times \frac{1}{6t}$

$\therefore -12y_2{t}^3=1+t^2$

$\therefore -36y_2{t}^3=3+3t^{2}from\left(1\right)and\left(2\right)weget,\Rightarrow 36y_2(y-\sqrt{x})=3+x$

On comparing above equation with the given equation we get,

$A=36$