Exam

JEE Main

Year

2025

Date

28 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Ellipse

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Ellipse

If the midpoint of a chord of the ellipse (x^2)/(9)+(y^2)/(4)=1 is...

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

Step-by-step solution

$$\begin{aligned} & E: \frac{x^2}{9}+\frac{y^2}{4}=1 \\ & T=S_1 \\ & \Rightarrow \frac{\sqrt{2} x}{9}+\frac{1}{4}\left(\frac{4}{3} y\right)-1=\frac{2}{9}+\frac{16}{9(4)}-1 \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9} \\ & \frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{3} \quad \Rightarrow \sqrt{2 x}+3 y=6 \end{aligned}$$ Now point of intersection of chord and ellipse is $$\begin{aligned} & \frac{(6-3 y)^2}{18}+\frac{y^2}{4}=1 \\ & \frac{(2-y)^2}{2}+\frac{y^2}{4}=1 \\ & 2\left(4+y^2-4 y\right)+y^2=4 \\ & \Rightarrow 3 y^2-8 y+4=0 \\ & \Rightarrow y=2, \frac{2}{3} \end{aligned}$$ 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} $$\begin{aligned} & =\sqrt{8+\frac{16}{9}} \\ & =\frac{\sqrt{88}}{3}=\frac{2 \sqrt{22}}{3} \end{aligned}$$$On comparing$\alpha=22