If `p_(1),p_(2),p_(3)` be the length of perpendiculars from the points `(m^(2),2m),(mm',m+m')` and `(m^('2),2m')` respectively on the line `xcosalpha+ysinalpha+(sin^(2)alpha)/(cosalpha)=0` then `p_(1),p_(2),p_(3)` are in:

Options

A. A.P
B. G.P
C. H.P
D. None of these

Answer

To solve the problem, we need to find the lengths of the perpendiculars from the given points to the specified line and determine the relationship between these lengths. Let's break it down step by step.

### Step 1: Identify the line and points
The line is given by the equation:
\[ x \cos \alpha + y \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha} = 0 \]

The points from which we need to find the perpendicular distances are:
1. \( P_1 = (m^2, 2m) \)
2. \( P_2 = (mm', m + m') \)
3. \( P_3 = (m'^2, 2m') \)

### Step 2: Use the formula for the length of the perpendicular
The length of the perpendicular \( p \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by:
\[ p = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]

For our line, we have:
- \( a = \cos \alpha \)
- \( b = \sin \alpha \)
- \( c = \frac{\sin^2 \alpha}{\cos \alpha} \)

### Step 3: Calculate \( P_1 \)
For the point \( (m^2, 2m) \):
\[ P_1 = \frac{|\cos \alpha \cdot m^2 + \sin \alpha \cdot 2m + \frac{\sin^2 \alpha}{\cos \alpha}|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} \]

Since \( \sqrt{\cos^2 \alpha + \sin^2 \alpha} = 1 \):
\[ P_1 = |\cos \alpha \cdot m^2 + 2m \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha}| \]

### Step 4: Simplify \( P_1 \)
Taking the common denominator:
\[ P_1 = \left| \frac{m^2 \cos^2 \alpha + 2m \sin \alpha \cos \alpha + \sin^2 \alpha}{\cos \alpha} \right| \]
This can be rewritten as:
\[ P_1 = \frac{|(m \cos \alpha + \sin \alpha)^2|}{\cos \alpha} \]

### Step 5: Calculate \( P_2 \)
For the point \( (mm', m + m') \):
\[ P_2 = \frac{|\cos \alpha \cdot mm' + \sin \alpha \cdot (m + m') + \frac{\sin^2 \alpha}{\cos \alpha}|}{1} \]
\[ P_2 = |\cos \alpha \cdot mm' + (m + m') \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha}| \]

Taking the common denominator:
\[ P_2 = \left| \frac{mm' \cos^2 \alpha + (m + m') \sin \alpha \cos \alpha + \sin^2 \alpha}{\cos \alpha} \right| \]
This can be rewritten as:
\[ P_2 = \frac{|(m \cos \alpha + \sin \alpha)(m' \cos \alpha + \sin \alpha)|}{\cos \alpha} \]

### Step 6: Calculate \( P_3 \)
For the point \( (m'^2, 2m') \):
\[ P_3 = \frac{|\cos \alpha \cdot m'^2 + 2m' \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha}|}{1} \]
Following the same steps as for \( P_1 \):
\[ P_3 = \frac{|(m' \cos \alpha + \sin \alpha)^2|}{\cos \alpha} \]

### Step 7: Establish the relationship between \( P_1, P_2, P_3 \)
From the expressions derived:
- \( P_1 = \frac{|(m \cos \alpha + \sin \alpha)^2|}{\cos \alpha} \)
- \( P_2 = \frac{|(m \cos \alpha + \sin \alpha)(m' \cos \alpha + \sin \alpha)|}{\cos \alpha} \)
- \( P_3 = \frac{|(m' \cos \alpha + \sin \alpha)^2|}{\cos \alpha} \)

### Step 8: Conclusion
We can see that:
\[ P_1 \cdot P_3 = P_2^2 \]
This indicates that \( P_1, P_2, P_3 \) are in geometric progression (GP).