Let `xtan alpha + ysin alpha= alpha` and `xalpha cosec alpha + ycosalpha= 1` be two variable straight lines, `alpha` being the parameter. Let `P` be the point of intersection of the lines. In the limiting position when `a ->0,` the point `P` lies on the line :

Options

A. `(2,-1)`
B. `(2,1)`
C. `(-2,1)`
D. `(-2,-1)`

Answer

To solve the problem step by step, we start with the two equations given in the question:

1. \( x \tan \alpha + y \sin \alpha = \alpha \) (Equation 1)
2. \( x \alpha \csc \alpha + y \cos \alpha = 1 \) (Equation 2)

We need to find the point \( P \) which is the intersection of these two lines as \( \alpha \) approaches 0.

### Step 1: Solve for \( x \) and \( y \)

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y \sin \alpha = \alpha - x \tan \alpha \]
\[ y = \frac{\alpha - x \tan \alpha}{\sin \alpha} \tag{3} \]

Now, substitute \( y \) from Equation 3 into Equation 2:
\[ x \alpha \csc \alpha + \left(\frac{\alpha - x \tan \alpha}{\sin \alpha}\right) \cos \alpha = 1 \]

### Step 2: Simplify Equation 2

Substituting \( y \) into Equation 2 gives:
\[ x \alpha \csc \alpha + \frac{(\alpha - x \tan \alpha) \cos \alpha}{\sin \alpha} = 1 \]
Multiplying through by \( \sin \alpha \):
\[ x \alpha + (\alpha - x \tan \alpha) \cos \alpha = \sin \alpha \]
Expanding this:
\[ x \alpha + \alpha \cos \alpha - x \tan \alpha \cos \alpha = \sin \alpha \]
Rearranging gives:
\[ x \alpha - x \tan \alpha \cos \alpha = \sin \alpha - \alpha \cos \alpha \]
Factoring out \( x \):
\[ x (\alpha - \tan \alpha \cos \alpha) = \sin \alpha - \alpha \cos \alpha \]
Thus,
\[ x = \frac{\sin \alpha - \alpha \cos \alpha}{\alpha - \tan \alpha \cos \alpha} \tag{4} \]

### Step 3: Find the limits as \( \alpha \to 0 \)

Now we need to find \( \lim_{\alpha \to 0} x \) and \( \lim_{\alpha \to 0} y \).

#### Finding \( \lim_{\alpha \to 0} x \)

Using L'Hôpital's Rule or Taylor series expansion:
\[ \tan \alpha \approx \alpha + \frac{\alpha^3}{3} + O(\alpha^5) \quad \text{and} \quad \sin \alpha \approx \alpha - \frac{\alpha^3}{6} + O(\alpha^5) \]
Thus,
\[ \sin \alpha - \alpha \cos \alpha \approx \alpha - \frac{\alpha^3}{6} - \alpha(1 - \frac{\alpha^2}{2}) = \frac{\alpha^3}{3} + O(\alpha^5) \]
And,
\[ \alpha - \tan \alpha \cos \alpha \approx \alpha - \left(\alpha + \frac{\alpha^3}{3}\right)(1 - \frac{\alpha^2}{2}) = \frac{\alpha^3}{6} + O(\alpha^5) \]
Thus,
\[ x \approx \frac{\frac{\alpha^3}{3}}{\frac{\alpha^3}{6}} = 2 \]

#### Finding \( \lim_{\alpha \to 0} y \)

Substituting \( x \) back into Equation 3:
\[ y = \frac{\alpha - 2 \tan \alpha}{\sin \alpha} \]
Using the Taylor expansion for \( \tan \alpha \):
\[ y \approx \frac{\alpha - 2\left(\alpha + \frac{\alpha^3}{3}\right)}{\alpha - \frac{\alpha^3}{6}} \approx \frac{-\alpha}{\alpha} = -1 \]

### Conclusion

Thus, the point \( P \) in the limiting position as \( \alpha \to 0 \) is:
\[ P = (2, -1) \]