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What are the points on the curve $ x^{3/2} + y^{3/2} = a^{3/2} $ where the tangents are equally inclined to the axes?

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What have you done, tried, achieved... in this question? – DonAntonio Jan 19 '13 at 17:51
Is this $a^{3/2}$ or $a^3 / 2$? I assumed it was the former, because you wrote "a^3{/2}", and the problem has thus far referred to $3/2$. – George V. Williams Jan 19 '13 at 17:51
@GeorgeV.Williams - sorry fixed now – AppDeveloper Jan 19 '13 at 17:52
till now i have tried to find the slope of the curve and equated to +-1 – AppDeveloper Jan 19 '13 at 17:53
I don't understand the question at all. Why tangents, in plural? The curve has only one tangent at each point! And inclined equally to what? And to what axis? – Harald Hanche-Olsen Jan 19 '13 at 17:57

$$x^{3/2}+y^{3/2}=\frac{a^3}{2}\Longrightarrow \frac{3}{2}\left(x^{1/2}\,dx+ y^{1/2}\,dy\right)=0\Longrightarrow$$

$$y'=-\sqrt\frac{x}{y}\Longrightarrow \,\text{two points}(x_1,y_1)\,,\,(x_2,y_2)\,\,\text{fulfill the condition}\Longleftrightarrow \frac{x_1}{x_2}=\frac{y_1}{y_2}$$

-- I'll leave it to you to check what happens when $\,x=0\,\,\vee\,\,\,y=0\,$ --

Since it must be that $\,x,y\geq 0\,$ (why?) , and also

$$y=\sqrt[3]{\left(\frac{a^3}{2}-x^{3/2}\right)^2}$$ we get

$$\frac{x_1}{x_2}=\frac{y_1}{y_2}=\left(\frac{\frac{a^3}{2}-x_1^{3/2}}{\frac{a^3}{2}-x_2^{3/2}}\right)^{2/3}\stackrel{\text{after a little algebra}}\Longleftrightarrow x_1^{3/2}=x_2^{3/2}$$


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The above answer is made with the original post where it was written $\,a^3/2\,$ and not the edited one of $\,a^{3/2}\,$ . Nevertheless, following the general lines of the above one can reach the answer to the new question. – DonAntonio Jan 19 '13 at 18:10

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