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← JEE Advanced 2026 · Paper 2
MathematicsNumerical+2Question 17 of 54

JEE Advanced 2026 · Paper 2, Mathematics question 17

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Question Stem for Question Nos. 17 and 18

Consider the ellipses given by
x2+4y2=1and4x2+y2=1.x^2+4y^2=1\qquad\text{and}\qquad4x^2+y^2=1.

Let PP be the point in the first quadrant where the given ellipses intersect. If θ\theta is the acute angle between the tangents to the given ellipses at the point PP, then the value of 4tan⁡θ4\tan\theta is ________.

Answer and worked solution

Official answer: 7.4

Subtracting the equations gives x2=y2x^2=y^2, so P=(15,15)P=\left(\frac1{\sqrt5},\frac1{\sqrt5}\right).

Slopes at PP. For x2+4y2=1x^2+4y^2=1: y′=−x4y=−14y'=-\frac{x}{4y}=-\frac14. For 4x2+y2=14x^2+y^2=1: y′=−4xy=−4y'=-\frac{4x}{y}=-4.
tan⁡θ=∣−14+41+1∣=158,4tan⁡θ=7.5.\tan\theta=\left|\frac{-\frac14+4}{1+1}\right|=\frac{15}8,\qquad4\tan\theta=7.5.

Topics:Tangents and normals to conicsAngle between lines
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