Skip to main content
Back to Conic Sections
JEE Main 2023
Conic Sections
Hyperbola
Easy

Question

A common tangent T\mathrm{T} to the curves C1:x24+y29=1\mathrm{C}_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 and C2:x242y2143=1C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1 does not pass through the fourth quadrant. If T\mathrm{T} touches C1\mathrm{C}_{1} at (x1,y1)\left(x_{1}, y_{1}\right) and C2\mathrm{C}_{2} at (x2,y2)\left(x_{2}, y_{2}\right), then 2x1+x2\left|2 x_{1}+x_{2}\right| is equal to ______________.

Answer: 2

Solution

This solution will guide you through finding the common tangent to an ellipse and a hyperbola, determining the correct tangent based on a quadrant condition, and then calculating the required expression involving the points of tangency.


1. Key Concepts: Equations of Tangents with Given Slope

A fundamental concept in conic sections is the ability to write the equation of a tangent line if its slope is known.

  • For an ellipse C1:x2a2+y2b2=1C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, the equation of a tangent line with slope mm is given by: y=mx±a2m2+b2y = mx \pm \sqrt{a^2m^2 + b^2}
  • For a hyperbola

Practice More Conic Sections Questions

View All Questions