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Ellipse

Section: Mathematics

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Question : 28 of 117

Marks: +1, -0

If the midpoint of a chord of the ellipse

\;\frac{x^2}{9}+\;\frac{y^2}{4}=1

\left(\sqrt{2},\;\frac{4}{3}\right)

, and the length of the chord is

\;\frac{2\sqrt{\alpha}}{3}

\alpha

[28 Jan 2025 Shift 2]

26
22
20
18

Solution: 👈: Video Solution

\text{E:}\;\frac{x^2}{9}+\;\frac{y^2}{4}=1

\;T=S_1

\;\Rightarrow\;\frac{\sqrt{2}x}{9}+\frac{1}{4}\left(\frac{4}{3}y\right)-1=\frac{2}{9}+\frac{16}{9\cdot4}-1

\;\frac{\sqrt{2}x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}

\;\frac{\sqrt{2}x}{9}+\frac{y}{3}=\frac{2}{3}\Rightarrow\sqrt{2x}+3y=6

Now point of intersection of chord and ellipse is

\;\frac{(6-3y)^2}{18}+\frac{y^2}{4}=1

\;\frac{(2-y)^2}{2}+\frac{y^2}{4}=1

\;2(4+y^2-4y)+y^2=4

\;\Rightarrow3y^2-8y+4=0

\;\Rightarrow y=2,\;\frac{2}{3}

So, points are

(0,2)

are

\left(2\sqrt{2},\;\frac{2}{3}\right)

Length of chord

=\sqrt{(2\sqrt{2})^2+\left(\frac{2}{3}-2\right)^2}

\;=\sqrt{8+\frac{16}{9}}

\;=\frac{\sqrt{88}}{3}=\frac{2\sqrt{22}}{3}

On comparing

\alpha=22

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