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Question Number 23428 by gopikrishnan005@gmail.com last updated on 30/Oct/17

$$thecurve{ay}^2=x^2(3a-x)cutsthey-axisat$$

Answered by $@ty@m last updated on 30/Oct/17

$$orogin$$

Answered by FilupES last updated on 31/Oct/17

$$y^2=\frac{x^2}{a}(3a-x)$$$$\mathrm{cuts}\mathrm{origin}\mathrm{when}y=0$$$$(\mathrm{note}{y}^2=y=0)$$$$$$$$\therefore \frac{x^2}{a}(3a-x)=0$$$$x=0$$$$\mathrm{and}x=3a$$

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{I}\mathrm{think}\mathrm{he}\mathrm{wants}\mathrm{the}\mathrm{cut}\mathrm{points}\mathrm{on}\mathrm{y}-\mathrm{axis},$$$$\mathrm{i}.\mathrm{e}.\mathrm{x}=0$$

Commented by FilupES last updated on 31/Oct/17

$$\mathrm{thanks},\mathrm{i}\mathrm{missread}\mathrm{the}\mathrm{question}$$

Commented by FilupES last updated on 31/Oct/17

$$yaxis\mathrm{when}x=0$$$$\therefore y=0$$

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