JEE Main 2025 Jan 28 Shift 2, Mathematics Q5: Ellipse: Chord, Tangent and Normal

JEE Main2025Jan 28, Shift 2Mathematics

If the midpoint of a chord of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is $(2, \frac{4}{3})$ , and the length of the chord is $\frac{2 α}{3}$ , then $α$ is :

A
$18$
B
$22$
C
$26$
D
$20$

Solution

Using the chord-with-given-midpoint relation $T = S_{1}$ on the ellipse with midpoint $M = (2, 4/3)$ :

$\frac{x 2}{9} + \frac{y}{4} \cdot \frac{4}{3} = \frac{2}{9} + \frac{16/9}{4} \Rightarrow \frac{2 x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{4}{9}$ .

Simplifying gives $2 x + 3 y = 6$ , i.e. $y = \frac{6 - 2 x}{3}$ .

Substituting into the ellipse and clearing denominators yields $6 x^{2} - 122 x = 0$ , so $x = 0$ or $x = 22$ , giving the endpoints $(0, 2)$ and $(22, 2/3)$ .

Chord length: $(22)^{2} + (2 - 2/3)^{2} = 8 + 16/9 = 88/9 = \frac{2 22}{3}$ .

$∴ α = 22$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q5: Ellipse: Chord, Tangent and Normal

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