JEE Advanced 2026 — Paper 2 — MATH-IV solution

JEE Advanced2026Paper 2MATH-IV

Question Stem (Q3-Q4).

Consider the ellipses given by

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

Q3. 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 ___________.

Your answer

Solution

Solving $x^{2} + 4 y^{2} = 1$ and $4 x^{2} + y^{2} = 1$ simultaneously: subtracting gives $3 x^{2} - 3 y^{2} = 0$ , so $x = \pm y$ . In the first quadrant, $x = y$ , and substituting back: $5 x^{2} = 1$ , hence $P = (\frac{1}{5}, \frac{1}{5})$ .

Slope at $P$ on ellipse 1: implicit differentiation of $x^{2} + 4 y^{2} = 1$ gives $2 x + 8 y y^{'} = 0$ , so $y^{'} = - \frac{x}{4 y}$ . At $P$ , $m_{1} = - \frac{1}{4}$ .

Slope at $P$ on ellipse 2: implicit differentiation of $4 x^{2} + y^{2} = 1$ gives $8 x + 2 y y^{'} = 0$ , so $y^{'} = - \frac{4 x}{y}$ . At $P$ , $m_{2} = - 4$ .

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

$∴ 4 tan θ = \frac{15}{2} = 7.50$ .

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