The locus of the point of intersection of tangents to the circle `x=a cos theta, y = a sin theta` at the points, whose parametric angles differ by `pi//2`, is

Options

A. straight line
B. circle
C. ellipse
D. none

Answer

To find the locus of the point of intersection of tangents to the circle defined by the parametric equations \( x = a \cos \theta \) and \( y = a \sin \theta \) at points whose parametric angles differ by \( \frac{\pi}{2} \), we can follow these steps:

### Step 1: Identify the points on the circle
Let the angle at point \( P_1 \) be \( \theta \) and at point \( P_2 \) be \( \theta + \frac{\pi}{2} \). The coordinates of these points on the circle are:
- \( P_1 = (a \cos \theta, a \sin \theta) \)
- \( P_2 = (a \cos(\theta + \frac{\pi}{2}), a \sin(\theta + \frac{\pi}{2}) ) \)

Using the trigonometric identities, we have:
- \( P_2 = (a \cos(\theta + \frac{\pi}{2}), a \sin(\theta + \frac{\pi}{2})) = (a \cdot 0, a \cdot 1) = (0, a) \)

### Step 2: Find the equations of the tangents at these points
The equation of the tangent to the circle at a point \( (x_0, y_0) \) is given by:
\[ xx_0 + yy_0 = a^2 \]
For point \( P_1 = (a \cos \theta, a \sin \theta) \):
\[ xx_1 + yy_1 = a^2 \implies x(a \cos \theta) + y(a \sin \theta) = a^2 \]
This simplifies to:
\[ x \cos \theta + y \sin \theta = a \]

For point \( P_2 = (0, a) \):
\[ x(0) + y(a) = a^2 \implies ay = a^2 \implies y = a \]

### Step 3: Find the point of intersection of the tangents
Now we have the two tangent equations:
1. \( x \cos \theta + y \sin \theta = a \)
2. \( y = a \)

Substituting \( y = a \) into the first equation:
\[ x \cos \theta + a \sin \theta = a \]
Rearranging gives:
\[ x \cos \theta = a - a \sin \theta \implies x = \frac{a(1 - \sin \theta)}{\cos \theta} \]

### Step 4: Eliminate \( \theta \) to find the locus
To find the locus, we can express \( x \) and \( y \) in terms of \( \theta \):
\[ y = a \quad \text{(constant)} \]
Substituting \( \sin \theta = \frac{y}{a} \) into the expression for \( x \):
\[ x = \frac{a(1 - \frac{y}{a})}{\sqrt{1 - \left(\frac{y}{a}\right)^2}} = \frac{a(1 - \frac{y}{a})}{\sqrt{\frac{a^2 - y^2}{a^2}}} = \frac{a(1 - \frac{y}{a}) a}{\sqrt{a^2 - y^2}} = \frac{a^2(1 - \frac{y}{a})}{\sqrt{a^2 - y^2}} \]

### Step 5: Identify the locus
The locus of the intersection points of the tangents is a circle. The general form of the locus is:
\[ x^2 + (y - a)^2 = a^2 \]
This represents a circle with center at \( (0, a) \) and radius \( a \).

### Final Answer
The locus of the point of intersection of tangents to the circle at points whose parametric angles differ by \( \frac{\pi}{2} \) is:
\[ x^2 + (y - a)^2 = a^2 \]