The locus of the point of intersection of tangents to the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` at the points whose eccentric angles differ by `pi//2`, is

Options

A. `x^(2) + y^(2) = a^(2)`
B. `x^(2) + y^(2) = b^(2)`
C. `x^(2) + y^(2) = a^(2) + b^(2)`
D. `x^(2)/a^(2) + y^(2)/b^(2) = 2`

Explanation

To find the locus of the point of intersection of tangents to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at points whose eccentric angles differ by \( \frac{\pi}{2} \), we can follow these steps:

### Step 1: Identify the Points on the Ellipse
Let the eccentric angle of the first point \( P \) be \( \theta \). The coordinates of point \( P \) on the ellipse are given by:
\[ P(a \cos \theta, b \sin \theta) \]
The second point \( Q \) will have an eccentric angle of \( \theta + \frac{\pi}{2} \). The coordinates of point \( Q \) are:
\[ Q\left(a \cos\left(\theta + \frac{\pi}{2}\right), b \sin\left(\theta + \frac{\pi}{2}\right)\right) \]
Using trigonometric identities, we find:
\[ \cos\left(\theta + \frac{\pi}{2}\right) = -\sin \theta \quad \text{and} \quad \sin\left(\theta + \frac{\pi}{2}\right) = \cos \theta \]
Thus, the coordinates of point \( Q \) become:
\[ Q(-a \sin \theta, b \cos \theta) \]

### Step 2: Write the Equations of the Tangents
The equation of the tangent to the ellipse at point \( P \) is given by:
\[ \frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \]
The equation of the tangent at point \( Q \) is:
\[ \frac{x}{-a \sin \theta} + \frac{y}{b \cos \theta} = 1 \]

### Step 3: Rearranging the Tangent Equations
Rearranging the first tangent equation gives:
\[ \frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \quad \Rightarrow \quad x + \frac{a \sin \theta}{b} y = a \cos \theta \]
Rearranging the second tangent equation gives:
\[ \frac{x}{-a \sin \theta} + \frac{y}{b \cos \theta} = 1 \quad \Rightarrow \quad -\frac{a \sin \theta}{b} y = x + a \sin \theta \]

### Step 4: Finding the Intersection Point
Let the point of intersection of these tangents be \( (h, k) \). Thus, we have:
1. \( h + \frac{a \sin \theta}{b} k = a \cos \theta \) (Equation 1)
2. \( -\frac{a \sin \theta}{b} k = h + a \sin \theta \) (Equation 2)

### Step 5: Solve the Equations
From Equation 1, we can express \( k \):
\[ k = \frac{b}{a \sin \theta}(a \cos \theta - h) \]
Substituting this into Equation 2 gives:
\[ -\frac{a \sin \theta}{b} \left(\frac{b}{a \sin \theta}(a \cos \theta - h)\right) = h + a \sin \theta \]
Simplifying this leads to:
\[ -(a \cos \theta - h) = h + a \sin \theta \]
Rearranging gives:
\[ 2h = a \cos \theta + a \sin \theta \]
Thus:
\[ h = \frac{a}{2} (\cos \theta + \sin \theta) \]

### Step 6: Find \( k \) in Terms of \( h \)
Substituting \( h \) back into the equation for \( k \):
\[ k = \frac{b}{a \sin \theta}\left(a \cos \theta - \frac{a}{2}(\cos \theta + \sin \theta)\right) \]
This simplifies to:
\[ k = \frac{b}{2 \sin \theta}(\cos \theta - \sin \theta) \]

### Step 7: Eliminate \( \theta \) to Find the Locus
Now we have \( h \) and \( k \) in terms of \( \theta \). To find the locus, we can square and add the equations for \( h \) and \( k \):
\[ \frac{h^2}{\left(\frac{a}{2}\right)^2} + \frac{k^2}{\left(\frac{b}{2}\right)^2} = 1 \]
This leads to:
\[ \frac{h^2}{\frac{a^2}{4}} + \frac{k^2}{\frac{b^2}{4}} = 1 \]
Multiplying through by 4 gives:
\[ \frac{h^2}{a^2} + \frac{k^2}{b^2} = 1 \]

### Final Result
Thus, the locus of the point of intersection of the tangents is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \]