Find number of tangent to given curve

Question

The number of tangent to curve $x^\frac{3}{2} +y^\frac{3}{2} = a^\frac{3}{2}$ where the tangents are equally inclined to axes, is

My work $$\frac{dy}{dx}=-\sqrt\frac{x}{y}$$ From above we can say that $$x,y>0$$ So x and y must lie in first coordinate . $$\frac{dy}{dx}=-\sqrt\frac{x}{y}=1$$ $$\sqrt\frac{x}{y}=-1$$ This is not possible $$\sqrt\frac{x}{y}=1$$ So there will be only one tangent$$ \sqrt x=\sqrt y$$ $$x=y$$ Please tell me whether I am correct or not.

Answer 1

Continuing from $\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}\tag{1}$

Yes you are correct, except that $x,y>0$ and $\sqrt{x}=\sqrt{y}$

Because when $x,y<0$, $\frac{x}{y} >0$ and $\sqrt\frac{x}{y}=1$ doesn't imply that $\sqrt{x}=\sqrt{y}$. For the sake of contradiction see that $\sqrt\frac{-2}{-2}=\sqrt{1}=1$, but $\sqrt{-2}=\sqrt{-2}$, is not defined, when $x,y$ are real.

So there is only one tangent in the first and third quadrant passing through the origin.

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Update:

But assumption that $(1)$ is correct is wrong because $$x^\frac{3}{2} +y^\frac{3}{2} = a^\frac{3}{2} \Rightarrow \dfrac{dx}{dy}=-\dfrac{\sqrt{x}}{\sqrt{y}} \text{ or } \dfrac{dx}{dy}=-\sqrt{\dfrac{x}{y}} \text{ when } x,y >0$$

So, $x,y>0$ and the required tangent is the first quadrant.

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Note:

There is nothing wrong with your solution except the above mentioned step(see Update). I just wanted to show you how a false step leads to other steps, the truth of which can't be determined, as logically it is true by default.

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Advice:

Just in case this:

$$\frac{dy}{dx}=-\sqrt\frac{x}{y}$$ From above we can say that $$x,y>0$$ So x and y must lie in first coordinate .

meant this:

$$ \dfrac{dx}{dy}=-\sqrt{\dfrac{x}{y}} \text{ when } x,y >0$$

I suggest that you use your words more carefully.