Consider the ellipses given by $$x^{2}+4 y^{2}=1 \text { and } 4 x^{2}+y^{2}=1 .$$ Let P be the point in the first quadrant where the given ellipses intersect. If $\theta$ is the acute angle between the tangents to the given ellipses at the point P , then the value of $4 \tan \theta$ is ____.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Ellipse

Consider the ellipses given by

x^{2} + 4 y^{2} = 1 and 4 x^{2} + y^{2} = 1.

Let P be the point in the first quadrant where the given ellipses intersect. If

θ

is the acute angle between the tangents to the given ellipses at the point P , then the value of

4 tan θ

is ____.

Answer: 7.5

For

x^{2} + 4 y^{2} = 1

, differentiating:

2 x + 8 y y^{'} = 0 \Rightarrow y^{'} = - \frac{x}{4 y}

. At intersection point P (found by solving the two ellipse equations), the slope is

m_{1} = - \frac{1}{4}

. For

4 x^{2} + y^{2} = 1

8 x + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{4 x}{y}

, giving

m_{2} = - 4

. The angle between tangents:

tan θ = ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣ = ∣ \frac{- \frac{1}{4} + 4}{1 + 1} ∣ = \frac{15/4}{2} = \frac{15}{8}

. Thus

4 tan θ = \frac{15}{2} = 7.50

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer