Points on curve where tangent are equally inclined
$$x^{3/2}+y^{3/2}=\frac{a^3}{2}\Longrightarrow \frac{3}{2}\left(x^{1/2}\,dx+ y^{1/2}\,dy\right)=0\Longrightarrow$$
$$y'=-\sqrt\frac{x}{y}\Longrightarrow \,\text{two points}(x_1,y_1)\,,\,(x_2,y_2)\,\,\text{fulfill the condition}\Longleftrightarrow \frac{x_1}{x_2}=\frac{y_1}{y_2}$$
-- 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
$$\frac{x_1}{x_2}=\frac{y_1}{y_2}=\left(\frac{\frac{a^3}{2}-x_1^{3/2}}{\frac{a^3}{2}-x_2^{3/2}}\right)^{2/3}\stackrel{\text{after a little algebra}}\Longleftrightarrow x_1^{3/2}=x_2^{3/2}$$
so...
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Parimal Raj
Passionate and enthusiastic developer with 6+ years of IT experience in development, implementation and testing of web applications using .NET framework, ASP MVC, React, Angular & NodeJS.
Updated on August 01, 2022Comments
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Parimal Raj over 1 year
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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DonAntonio almost 11 yearsWhat have you done, tried, achieved... in this question?
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George V. Williams almost 11 yearsIs 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 $3/2$.
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Parimal Raj almost 11 years@GeorgeV.Williams - sorry fixed now
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Parimal Raj almost 11 yearstill now i have tried to find the slope of the curve and equated to +-1
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Harald Hanche-Olsen almost 11 yearsI 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?
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Parimal Raj almost 11 years@HaraldHanche-Olsen - edited for grammer change!
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George V. Williams almost 11 yearsIs the question to find when the derivative of the function is $1$ or $-1$?
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DonAntonio almost 11 yearsThe 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.