The number of tangents to the curve `x^(2//3)+y^(2//3)=a^(2//3)` which are equally inclined to the axes is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To find the number of tangents to the curve \( x^{2/3} + y^{2/3} = a^{2/3} \) that are equally inclined to the axes, we can follow these steps:

### Step 1: Differentiate the equation
We start with the equation of the curve:
\[ x^{2/3} + y^{2/3} = a^{2/3} \]
To find the slope of the tangents, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = 0 \]
Using the chain rule, we have:
\[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the differentiated equation gives:
\[ \frac{2}{3}y^{-1/3} \frac{dy}{dx} = -\frac{2}{3}x^{-1/3} \]
Dividing both sides by \(\frac{2}{3}\):
\[ y^{-1/3} \frac{dy}{dx} = -x^{-1/3} \]
Now, multiplying both sides by \( y^{1/3} \):
\[ \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} \]

### Step 3: Set the slope equal to 1 and -1
Since we are looking for tangents that are equally inclined to the axes, we set:
\[ \frac{dy}{dx} = 1 \quad \text{and} \quad \frac{dy}{dx} = -1 \]

### Step 4: Solve for \( y \) in terms of \( x \)
1. For \(\frac{dy}{dx} = 1\):
\[ -\frac{y^{1/3}}{x^{1/3}} = 1 \implies y^{1/3} = -x^{1/3} \implies y = -x \]

2. For \(\frac{dy}{dx} = -1\):
\[ -\frac{y^{1/3}}{x^{1/3}} = -1 \implies y^{1/3} = x^{1/3} \implies y = x \]

### Step 5: Substitute \( y = x \) and \( y = -x \) into the original equation
1. Substituting \( y = x \):
\[ x^{2/3} + x^{2/3} = a^{2/3} \implies 2x^{2/3} = a^{2/3} \implies x^{2/3} = \frac{a^{2/3}}{2} \]
Taking the cube:
\[ x^2 = \frac{a^2}{4} \implies x = \pm \frac{a}{2} \]
Thus, the points are:
\[ \left(\frac{a}{2}, \frac{a}{2}\right) \quad \text{and} \quad \left(-\frac{a}{2}, -\frac{a}{2}\right) \]

2. Substituting \( y = -x \):
\[ x^{2/3} + (-x)^{2/3} = a^{2/3} \implies 2x^{2/3} = a^{2/3} \implies x^{2/3} = \frac{a^{2/3}}{2} \]
The same calculations yield:
\[ x = \pm \frac{a}{2} \]
Thus, the points are:
\[ \left(\frac{a}{2}, -\frac{a}{2}\right) \quad \text{and} \quad \left(-\frac{a}{2}, \frac{a}{2}\right) \]

### Step 6: Count the tangents
We have found four points of tangency:
1. \(\left(\frac{a}{2}, \frac{a}{2}\right)\)
2. \(\left(-\frac{a}{2}, -\frac{a}{2}\right)\)
3. \(\left(\frac{a}{2}, -\frac{a}{2}\right)\)
4. \(\left(-\frac{a}{2}, \frac{a}{2}\right)\)

Thus, the total number of tangents that are equally inclined to the axes is:
\[ \boxed{4} \]