Question

An angle of intersection of the curves, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) and \( x^2 + y^2 = ab, a>b \), is :

• A. \( \tan^{-1}\left( \frac{a - b}{\sqrt{ab}} \right) \)
• B. \( \tan^{-1}\left( \frac{a + b}{\sqrt{ab}} \right) \)
• C. \( \tan^{-1}\left( \frac{a - b}{2\sqrt{ab}} \right) \)
• D. \( \tan^{-1}(2\sqrt{ab}) \)

Hint:

For symmetrical curves like ellipse and circle, the angle of intersection is the same at all four intersection points.
Using homogenous coordinates or specific substitutions for \( x^2, y^2 \) simplifies the algebra.

Answer 1

The Correct Option is A

To find the angle of intersection between the curves \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 and x^2 + y^2 = ab , we can make use of the formula for the angle between two curves at a point of intersection. The angle \theta between the two curves is given by:

\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

where m_1 and m_2 are the slopes of the tangents to the given curves at the point of intersection.

Let's find the expressions for m_1 and m_2 :

1. For the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, the slope of the tangent line is obtained by differentiating implicitly with respect to x:

\frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right) = 0

This gives us:

\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2}{a^2} \frac{x}{y}

Thus, m_1 = -\frac{b^2}{a^2} \frac{x}{y} .

2. For the circle x^2 + y^2 = ab, the slope of the tangent line is:

\frac{d}{dx}(x^2 + y^2) = 0

This differentiates to:

2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

Thus, m_2 = -\frac{x}{y} .

Now, substitute m_1 and m_2 into the formula for the tangent of the angle \theta:

\tan \theta = \left| \frac{-\frac{b^2}{a^2} \frac{x}{y} + \frac{x}{y}}{1 + \left(-\frac{b^2}{a^2} \frac{x}{y}\right) \left(-\frac{x}{y}\right)} \right| \implies \tan \theta = \left| \frac{\left(-\frac{b^2}{a^2} + 1\right)\frac{x}{y}}{1 + \frac{b^2}{a^2} \frac{x^2}{y^2}} \right|

Now, let's evaluate at the intersection of the ellipse and the circle. Since both curves satisfy x^2 + y^2 = ab, substitute y^2 = ab - x^2 back into the angle expression to find the exact \theta:

After simplifying, we will find:

\theta = \tan^{-1} \left( \frac{a - b}{\sqrt{ab}} \right)

Therefore, the correct option is:

\( \tan^{-1}\left( \frac{a - b}{\sqrt{ab}} \right) \)