JEE Challenger
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MathMedAdv 2026 • Paper 2

Calculate Angle Between Tangents of Intersecting Ellipses

Comprehension Passage

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

Official Numerical Answer7.4 to 7.6

Topics & Concepts

Conic SectionsEllipseTangents and Normals

Step-by-Step Solution

To find the required value, we first determine the point of intersection PP of the two given ellipses x2+4y2=1x^2 + 4y^2 = 1 and 4x2+y2=14x^2 + y^2 = 1 in the first quadrant, which yields P=(15,15)P = \left(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right).

Next, we evaluate the slopes of the tangents to each ellipse at point PP using implicit differentiation: m1=−x4y=−14andm2=−4xy=−4m_1 = -\frac{x}{4y} = -\frac{1}{4} \quad \text{and} \quad m_2 = -\frac{4x}{y} = -4

Using the formula for the acute angle θ\theta between two intersecting lines with slopes m1m_1 and m2m_2: tan⁡θ=∣m1−m21+m1m2∣=∣−14−(−4)1+(−14)(−4)∣=158\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\frac{1}{4} - (-4)}{1 + \left(-\frac{1}{4}\right)(-4)} \right| = \frac{15}{8}

Finally, we compute the requested value: 4tan⁡θ=4×158=152=7.54 \tan \theta = 4 \times \frac{15}{8} = \frac{15}{2} = 7.5

Calculate Angle Between Tangents of Intersecting Ellipses | Mathematics PYQ Solution - JEE Challenger