Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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

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

-- 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...

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.