The length of the chord of the ellipse 4 x 2 + 2 y 2 = 1 , whose mid-point is ( 1 , 2 1 ) , is :

T = S 1 4 x ⋅ 1 + 4 y = 4 1 + 8 1 ⇒ 2 x + 2 y = 3 4 x 2 + 2 ( 2 3 − 2 x ) 2 = 1 ⇒ x = 12 12 ± 120 ⇒ y = 2 1 ∓ 12 120 So length of chord = 3 2 15

Mathematics - Ellipse Question with Solution | TestHub

MathematicsEllipseGeneralMedium2 minPYQ_2025

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The length of the chord of the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ , whose mid-point is $(1, \frac{1}{2})$ , is :

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Answer:

Solution:

$T = S_{1} \frac{x \cdot 1}{4} + \frac{y}{4} = \frac{1}{4} + \frac{1}{8} \Rightarrow 2 x + 2 y = 3 \frac{x ^{2}}{4} + \frac{( \frac{3 - 2 x}{2} ) ^{2}}{2} = 1 \Rightarrow x = \frac{12 \pm 120}{12} \Rightarrow y = \frac{1}{2} \mp \frac{120}{12}$

So length of chord
$= \frac{2 15}{3}$

Stream:JEESubject:MathematicsTopic:EllipseSubtopic:General

⏱ 2mℹ️ Source: PYQ_2025

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