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Question Number 45112 by ajfour last updated on 09/Oct/18

Commented by MrW3 last updated on 09/Oct/18

$$allcorrectsir!$$

Commented by ajfour last updated on 09/Oct/18

$$y=b\sin \alpha [\cosh \left(\frac{2x}{b\sin 2\alpha }\right)+\sin \alpha \sinh \left(\frac{2x}{b\sin 2\alpha }\right)-1]$$$$forb=3,\alpha =\frac{\pi }{4};wehavethe$$$$graphabove.$$

Commented by ajfour last updated on 09/Oct/18

$$relatedtoQ.45082\mathrm{\&}Q.44872$$

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