Angle between the curve `x^2/a^2+y^2/b^2=1` and `x^2+y^2=ab,a gt b`.

Answer

To find the angle between the curves given by the equations \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (an ellipse) and \( x^2 + y^2 = ab \) (a circle), we will follow these steps:

### Step 1: Find the intersection points of the curves
We start with the equations of the curves:
1. Ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
2. Circle: \( x^2 + y^2 = ab \)

To find the intersection points, we can express \( x^2 \) from the circle's equation:
\[ x^2 = ab - y^2 \]
Now, substitute this expression into the ellipse's equation:
\[ \frac{ab - y^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Multiply through by \( a^2b^2 \) to eliminate the denominators:
\[ b^2(ab - y^2) + a^2y^2 = a^2b^2 \]
This simplifies to:
\[ ab^3 - b^2y^2 + a^2y^2 = a^2b^2 \]
Rearranging gives:
\[ y^2(a^2 - b^2) = a^2b^2 - ab^3 \]
Thus,
\[ y^2 = \frac{a^2b^2 - ab^3}{a^2 - b^2} \]

### Step 2: Calculate the y-coordinates of the intersection points
Substituting the expression for \( y^2 \) back into the circle's equation gives:
\[ x^2 = ab - \frac{a^2b^2 - ab^3}{a^2 - b^2} \]
After simplification, we find:
\[ x^2 = \frac{a^2b + ab^2}{a^2 - b^2} \]

### Step 3: Find the slopes of the tangents at the intersection points
To find the slopes of the tangents at the intersection points, we differentiate both equations.

1. For the circle \( x^2 + y^2 = ab \):
\[ 2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \quad (m_1) \]

2. For the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \):
\[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2x}{a^2y} \quad (m_2) \]

### Step 4: Use the angle formula
The angle \( \theta \) between the two curves can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{-\frac{x}{y} + \frac{b^2x}{a^2y}}{1 - \frac{b^2x^2}{a^2y^2}} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{x(b^2 - a^2)}{a^2y^2 + b^2x^2} \right| \]

### Step 5: Substitute the intersection points into the angle formula
Now, substituting the values of \( x \) and \( y \) from the intersection points into the expression for \( \tan \theta \) gives the final angle:
\[ \theta = \tan^{-1} \left( \frac{a - b}{\sqrt{ab}} \right) \]

### Final Answer
Thus, the angle between the two curves is:
\[ \theta = \tan^{-1} \left( \frac{a - b}{\sqrt{ab}} \right) \]