If $x=a\cos t,y=a\sin t\text{ then }\frac{d^{3}y}{d{x}^3}\text{ at }t=\frac{\pi }{4}$ is $k/a^2$ then k=

Clearly $x^2+y^2=a^2\text{ and }y(\pi /4)=a/\sqrt{2},x(\pi /4)$

$=a/\sqrt{2}.$ Differentiating we get

$2x+2y{y}_1=0\Rightarrow y_1=-\frac{x}{y}\text{, so }{y}_1(\pi /4)=-1\text{. }$

Now $x+y{y}_1=0\Rightarrow 1+y_1^2+y{y}_2=0$

$\Rightarrow y_2(\pi /4)=-\frac{1+{\left(y_1(\pi /4)\right)}^2}{y(\pi /4)}=\frac{-2\sqrt{2}}{a}$

Differentiating again, we have

$2y_1{y}_2+y_1{y}_2+y{y}_3=0\Rightarrow y_3=-\frac{3y_1{y}_2}{y}$

$\Rightarrow y_3(\pi /4)=-\frac{3(-1)(-2)\sqrt{2}}{a\cdot a/\sqrt{2}}=-\frac{12}{a^2}.$