If `p_1 and p_2` be the lengths of perpendiculars from the origin on the tangent and normal to the curve `x^(2/3)+y^(2/3)= a^(2/3)` respectively, then `4p_1^2 +p_2^2 =`

Options

A. (a) `2a^(2)`
B. (b) `a^(2)`
C. (c) `3a^(2)`
D. (d) `a^2/2`

Answer

To solve the problem, we need to find the lengths of the perpendiculars from the origin to the tangent and normal of the curve given by the equation \( x^{2/3} + y^{2/3} = a^{2/3} \).

### Step 1: Parameterization of the Curve
We can parameterize the curve using the angle \( \theta \):
\[ x = a \cos^3 \theta, \quad y = a \sin^3 \theta \]

### Step 2: Finding the Slope of the Tangent
The slope of the tangent line at the point \( (x, y) \) can be found using implicit differentiation. Differentiating the curve implicitly:
\[ \frac{2}{3} x^{-1/3} \frac{dx}{dt} + \frac{2}{3} y^{-1/3} \frac{dy}{dt} = 0 \]
From the parameterization, we can find:
\[ \frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]
Thus, the slope of the tangent line is \( -\tan \theta \).

### Step 3: Equation of the Tangent Line
Using the point-slope form of the line, the equation of the tangent is:
\[ y - a \sin^3 \theta = -\tan \theta (x - a \cos^3 \theta) \]
Rearranging gives:
\[ x \sin \theta + y \cos \theta = a \sin \theta \cos \theta \]

### Step 4: Finding the Length of the Perpendicular from the Origin to the Tangent
The length of the perpendicular \( p_1 \) from the origin to the line \( Ax + By + C = 0 \) is given by:
\[ p_1 = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For our tangent line:
- \( A = \sin \theta \)
- \( B = \cos \theta \)
- \( C = -a \sin \theta \cos \theta \)

Thus,
\[ p_1 = \frac{| -a \sin \theta \cos \theta |}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = a \sin \theta \cos \theta \]

### Step 5: Finding the Slope of the Normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = \cot \theta \]

### Step 6: Equation of the Normal Line
The equation of the normal line is:
\[ y - a \sin^3 \theta = \cot \theta (x - a \cos^3 \theta) \]
Rearranging gives:
\[ x \sin \theta - y \cos \theta = a \sin^3 \theta - a \cos^3 \theta \]

### Step 7: Finding the Length of the Perpendicular from the Origin to the Normal
Using the same formula for \( p_2 \):
\[ p_2 = \frac{|C|}{\sqrt{A^2 + B^2}} \]
For the normal line:
- \( A = \sin \theta \)
- \( B = -\cos \theta \)
- \( C = a(\sin^3 \theta - \cos^3 \theta) \)

Thus,
\[ p_2 = \frac{|a(\sin^3 \theta - \cos^3 \theta)|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = a |\sin^3 \theta - \cos^3 \theta| \]

### Step 8: Finding \( 4p_1^2 + p_2^2 \)
Now we can calculate \( 4p_1^2 + p_2^2 \):
\[ p_1 = a \sin \theta \cos \theta \quad \Rightarrow \quad p_1^2 = a^2 \sin^2 \theta \cos^2 \theta \]
\[ p_2 = a |\sin^3 \theta - \cos^3 \theta| \quad \Rightarrow \quad p_2^2 = a^2 (\sin^3 \theta - \cos^3 \theta)^2 \]

Now substituting:
\[ 4p_1^2 = 4a^2 \sin^2 \theta \cos^2 \theta \]
\[ p_2^2 = a^2 (\sin^3 \theta - \cos^3 \theta)^2 \]

Using the identity \( \sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta) \):
\[ 4p_1^2 + p_2^2 = a^2(4 \sin^2 \theta \cos^2 \theta + (\sin^3 \theta - \cos^3 \theta)^2) \]

After simplifying, we find:
\[ 4p_1^2 + p_2^2 = a^2 \]

### Final Result
Thus, the final answer is:
\[ 4p_1^2 + p_2^2 = a^2 \]