JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let $(h, k)$ lie on the circle $C : x^{2} + y^{2} = 4$ and the point $(2 h + 1, 3 k + 2)$ lie on an ellipse with eccentricity $e$ . Then the value of $\frac{5}{e ^{2}}$ is equal to ________.

Your answer

Solution

Parametrise the circle: $(h, k) = (2 cos θ, 2 sin θ)$ .

Then $(2 h + 1, 3 k + 2) = (4 cos θ + 1, 6 sin θ + 2)$ .

Letting $X = 4 cos θ + 1$ and $Y = 6 sin θ + 2$ :

$(\frac{X - 1}{4})^{2} + (\frac{Y - 2}{6})^{2} = 1$ .

This is an ellipse with semi-axes 4 and 6. Since $6 > 4$ , the major axis is along $Y$ and

$e^{2} = 1 - \frac{4 ^{2}}{6 ^{2}} = 1 - \frac{16}{36} = \frac{20}{36} = \frac{5}{9}$ .

$\frac{5}{e ^{2}} = \frac{5}{5/9} = 9$ .

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