If the mid-point of a chord of the ellipse is , and the length of the chord is , then is: [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If the mid-point of a chord of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is $(\sqrt{2}, \frac{4}{3})$ and the length of the chord is $\frac{2 \sqrt{α}}{3}$ , then $α$ is: [2025]

1 22

2 26

3 20

4 18

Ans.
(1)

Let AB is a chord and M is the mid-point.

If $M (\sqrt{2}, \frac{4}{3})$ then equation of AB is

$T = S_{1} \Rightarrow \frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} - 1 = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} - 1$

$\Rightarrow \frac{x \sqrt{2}}{9} + \frac{y}{4} (\frac{4}{3}) = \frac{{(\sqrt{2})}^{2}}{9} + \frac{{(\frac{4}{3})}^{2}}{4}$

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

$\Rightarrow \sqrt{2} x + 3 y = 6 \Rightarrow y = \frac{6 - \sqrt{2} x}{3}$

Putting in ellipse, we get $\frac{x^{2}}{9} + \frac{{(6 - \sqrt{2} x)}^{2}}{9 \times 4} = 1$

$\Rightarrow 4 x^{2} + 36 + 2 x^{2} - 12 \sqrt{2} x = 36$

$\Rightarrow 6 x^{2} - 12 \sqrt{2} x = 0 \Rightarrow 6 x (x - 2 \sqrt{2}) = 0$

$\Rightarrow x = 0 and x = 2 \sqrt{2}$

So, y = 2 and $y = \frac{2}{3}$

$∴$ Length of the chord = $= \sqrt{{(2 \sqrt{2} - 0)}^{2} + {(\frac{2}{3} - 2)}^{2}}$

$= \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2}{3} \sqrt{22}$

So, $α = 22$ .

Want Us to Email you About Special Offers & Updates?