A curve is given by the equations $x=a\cos heta +\frac{1}{2}b\cos 2 heta ,$ $y=a\sin heta +\frac{1}{2}b\,\sin \,2 heta $, then the points for which $\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,$ is given by

Question

A curve is given by the equations $x=a\cos 	heta +\frac{1}{2}b\cos 2	heta ,$ $y=a\sin 	heta +\frac{1}{2}b\,\sin \,2	heta $, then the points for which $\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,$ is given by

Accepted Answer

$\cos 	heta =\frac{-\left( {{a}^{2}}+2{{b}^{2}} ight)}{3ab}$

Answer

$\sin 	heta =\frac{2{{a}^{2}}+{{b}^{2}}}{5ab}$

Answer

$	an 	heta =\frac{3{{a}^{2}}+2{{b}^{2}}}{4ab}$

Answer

$\cos 	heta =\frac{\left( {{a}^{2}}-2{{b}^{2}} ight)}{3ab}$

differentiation

Question: A curve is given by the equations $x=a\cos \theta +\frac{1}{2}b\cos 2\theta ,$ $y=a\sin \theta +\frac{1}{2}b\,\sin \,2\theta $, then the points for which $\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,$ is given by

differentiation

Question: A curve is given by the equations $x=a\cos \theta +\frac{1}{2}b\cos 2\theta ,$ $y=a\sin \theta +\frac{1}{2}b\,\sin \,2\theta $, then the points for which $\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,$ is given by

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