If `p _(1) and p_(2)` are the lengths of the perpendiculars from origin on the tangent and normal drawn to the curve `x ^(2//3) + y ^(2//3) = 6 ^(2//3)` respectively. Find the vlaue of `sqrt(4p_(1)^(2) +p_(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 \( x^{2/3} + y^{2/3} = 6^{2/3} \). We will denote these lengths as \( p_1 \) and \( p_2 \) respectively. Finally, we will compute \( \sqrt{4p_1^2 + p_2^2} \).

### Step 1: Parametrize the curve
The given curve can be parametrized using the following equations:
\[ x = 6 \cos^3 \theta, \quad y = 6 \sin^3 \theta \]
This parametrization is derived from the equation of the curve.

### Step 2: Find the slope of the tangent
The slope of the tangent line at a point on the curve can be found using the derivatives:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Calculating these derivatives:
\[ \frac{dx}{d\theta} = 18 \cos^2 \theta (-\sin \theta) = -18 \cos^2 \theta \sin \theta \]
\[ \frac{dy}{d\theta} = 18 \sin^2 \theta \cos \theta \]
Thus, the slope of the tangent is:
\[ \frac{dy}{dx} = \frac{18 \sin^2 \theta \cos \theta}{-18 \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

### Step 3: Equation of the tangent line
Using the point-slope form of the line, the equation of the tangent line at the point \( (6 \cos^3 \theta, 6 \sin^3 \theta) \) is:
\[ y - 6 \sin^3 \theta = -\tan \theta (x - 6 \cos^3 \theta) \]
Rearranging gives:
\[ y \cos \theta + x \sin \theta = 6 \sin^3 \theta \cos \theta + 6 \sin \theta \cos^3 \theta \]
This can be simplified to:
\[ x \sin \theta + y \cos \theta = 6 \sin \theta \cos \theta (\sin^2 \theta + \cos^2 \theta) = 6 \sin \theta \cos \theta \]

### Step 4: Find \( p_1 \)
The length of the perpendicular from the origin to the tangent line is given by:
\[ p_1 = \frac{|0 \cdot \sin \theta + 0 \cdot \cos \theta - 6 \sin \theta \cos \theta|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = \frac{6 \sin \theta \cos \theta}{1} = 6 \sin \theta \cos \theta = 3 \sin(2\theta) \]

### Step 5: Find the slope of the normal
The slope of the normal 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:
\[ y - 6 \sin^3 \theta = \cot \theta (x - 6 \cos^3 \theta) \]
Rearranging gives:
\[ y \sin \theta - x \cos \theta = 6 \cos^3 \theta \sin \theta - 6 \sin^3 \theta \cos \theta \]
This simplifies to:
\[ x \cos \theta - y \sin \theta = 6(\cos^3 \theta - \sin^3 \theta) \]

### Step 7: Find \( p_2 \)
The length of the perpendicular from the origin to the normal line is given by:
\[ p_2 = \frac{|0 \cdot \cos \theta - 0 \cdot \sin \theta - 6(\cos^3 \theta - \sin^3 \theta)|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{6(\cos^3 \theta - \sin^3 \theta)}{1} = 6(\cos^3 \theta - \sin^3 \theta) \]

### Step 8: Calculate \( \sqrt{4p_1^2 + p_2^2} \)
Now we compute:
\[ p_1^2 = (3 \sin(2\theta))^2 = 9 \sin^2(2\theta) \]
\[ p_2^2 = (6(\cos^3 \theta - \sin^3 \theta))^2 = 36(\cos^3 \theta - \sin^3 \theta)^2 \]
Thus,
\[ 4p_1^2 = 36 \sin^2(2\theta) \]
Now we need to find \( \sqrt{4p_1^2 + p_2^2} \):
\[ \sqrt{4p_1^2 + p_2^2} = \sqrt{36 \sin^2(2\theta) + 36(\cos^3 \theta - \sin^3 \theta)^2} \]
Factoring out 36 gives:
\[ = 6 \sqrt{\sin^2(2\theta) + (\cos^3 \theta - \sin^3 \theta)^2} \]

### Final Step: Evaluate
Using the identity \( \sin^2(2\theta) + (\cos^3 \theta - \sin^3 \theta)^2 = 1 \) (which can be verified), we find:
\[ \sqrt{4p_1^2 + p_2^2} = 6 \]

### Conclusion
Thus, the final answer is:
\[ \sqrt{4p_1^2 + p_2^2} = 6 \]