If `pi+theta` is the eccentric angle of a point on the ellipse `16x^(2)+25y^(2) = 400` then the corresponding point on the auxiliary circle is

Options

A. `(-4 cos theta,-4 sin theta)`
B. `(-5 cos theta,-5 sin theta)`
C. `(4 cos theta,4 sin theta)`
D. `(5 cos theta,5 sin theta)`

Answer

To solve the problem, we need to find the corresponding point on the auxiliary circle for a given eccentric angle on the ellipse defined by the equation \( 16x^2 + 25y^2 = 400 \).

### Step 1: Identify the semi-major and semi-minor axes of the ellipse.
The given equation of the ellipse can be rewritten in standard form:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
From this, we can identify the semi-major axis \( a \) and the semi-minor axis \( b \):
- \( a = 5 \) (since \( a^2 = 25 \))
- \( b = 4 \) (since \( b^2 = 16 \))

### Step 2: Determine the eccentric angle.
The eccentric angle is given as \( \pi + \theta \).

### Step 3: Find the coordinates of the point on the ellipse.
Using the parametric equations for the ellipse, the coordinates of a point on the ellipse corresponding to the eccentric angle \( \phi \) are given by:
\[ (x, y) = (a \cos \phi, b \sin \phi) \]
Substituting \( a \) and \( b \) and \( \phi = \pi + \theta \):
\[ x = 5 \cos(\pi + \theta) = 5(-\cos \theta) = -5 \cos \theta \]
\[ y = 4 \sin(\pi + \theta) = 4(-\sin \theta) = -4 \sin \theta \]
Thus, the coordinates of the point on the ellipse are:
\[ (-5 \cos \theta, -4 \sin \theta) \]

### Step 4: Find the corresponding point on the auxiliary circle.
The auxiliary circle of the ellipse has a radius equal to the semi-major axis \( a = 5 \). The coordinates of a point on the auxiliary circle corresponding to the eccentric angle \( \phi \) are given by:
\[ (x', y') = (a \cos \phi, a \sin \phi) \]
Substituting \( a = 5 \) and \( \phi = \pi + \theta \):
\[ x' = 5 \cos(\pi + \theta) = 5(-\cos \theta) = -5 \cos \theta \]
\[ y' = 5 \sin(\pi + \theta) = 5(-\sin \theta) = -5 \sin \theta \]
Thus, the coordinates of the point on the auxiliary circle are:
\[ (-5 \cos \theta, -5 \sin \theta) \]

### Final Answer:
The corresponding point on the auxiliary circle is:
\[ (-5 \cos \theta, -5 \sin \theta) \]