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JEE Main 2023
Conic Sections
Ellipse
Hard

Question

Let the ellipse 3x2+py2=43x^2 + py^2 = 4 pass through the centre CC of the circle x2+y22x4y11=0x^2 + y^2 - 2x - 4y - 11 = 0 of radius rr. Let f1,f2f_1, f_2 be the focal distances of the point CC on the ellipse. Then 6f1f2r6f_1f_2 - r is equal to

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Solution

This problem requires us to combine concepts from circles and ellipses, specifically finding the center and radius of a circle, determining the equation and properties of an ellipse, and calculating focal distances.


1. Key Concepts and Formulas

  • Circle Equation: The general equation of a circle is x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0. Its center is C(g,f)C(-g, -f) and its radius is r=g2+f2cr = \sqrt{g^2+f^2-c}. Alternatively, the standard form is (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2, where C(h,k)C(h,k) is the center and rr is the radius.

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