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JEE Main 2019
Conic Sections
Ellipse
Medium

Question

An angle of intersection of the curves, x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1 and x 2 + y 2 = ab, a > b, is :

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Solution

1. Understanding the Problem and Key Concept

The problem asks us to find the angle of intersection between two given curves:

  • Curve 1 (Ellipse): C1:x2a2+y2b2=1C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
  • Curve 2 (Circle): C2:x2+y2=abC_2: x^2 + y^2 = ab We are given that a>ba > b.

Key Concept: The angle between two intersecting curves at a point (x0,y0)(x_0, y_0) is defined as the angle between their respective tangent lines at that point. If m1m_1 and m2m_2 are the slopes of the tangent lines to the two curves at their

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