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Question Number 23428 by gopikrishnan005@gmail.com last updated on 30/Oct/17
the curve ay^2 =x^2 (3a−x) cuts the y-axis at
$${the}\:{curve}\:{ay}^{\mathrm{2}} ={x}^{\mathrm{2}} \left(\mathrm{3}{a}−{x}\right)\:{cuts}\:{the}\:{y}-{axis}\:{at} \\ $$
Answered by $@ty@m last updated on 30/Oct/17
orogin
$${orogin} \\ $$
Answered by FilupES last updated on 31/Oct/17
y^2 =(x^2 /a)(3a−x)  cuts origin when y=0  (note y^2 =y=0)     ∴(x^2 /a)(3a−x)=0  x=0  and x=3a
$${y}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{{a}}\left(\mathrm{3}{a}−{x}\right) \\ $$$$\mathrm{cuts}\:\mathrm{origin}\:\mathrm{when}\:{y}=\mathrm{0} \\ $$$$\left(\mathrm{note}\:{y}^{\mathrm{2}} ={y}=\mathrm{0}\right) \\ $$$$\: \\ $$$$\therefore\frac{{x}^{\mathrm{2}} }{{a}}\left(\mathrm{3}{a}−{x}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$$\mathrm{and}\:{x}=\mathrm{3}{a} \\ $$
Commented by mrW1 last updated on 31/Oct/17
I think he wants the cut points on y−axis,  i.e. x=0
$$\mathrm{I}\:\mathrm{think}\:\mathrm{he}\:\mathrm{wants}\:\mathrm{the}\:\mathrm{cut}\:\mathrm{points}\:\mathrm{on}\:\mathrm{y}−\mathrm{axis}, \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{x}=\mathrm{0} \\ $$
Commented by FilupES last updated on 31/Oct/17
thanks, i missread the question
$$\mathrm{thanks},\:\mathrm{i}\:\mathrm{missread}\:\mathrm{the}\:\mathrm{question} \\ $$
Commented by FilupES last updated on 31/Oct/17
y axis when x=0  ∴y=0
$${y}\:{axis}\:\mathrm{when}\:{x}=\mathrm{0} \\ $$$$\therefore{y}=\mathrm{0} \\ $$

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