The length of the chord of the ellipse , whose mid-point is , is: [2025]

Question

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: [2025]

Smart Achievers · Accepted Answer

(3)

We have, $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$ ,,, (i)

Mid-point of chord is $(1, \frac{1}{2})$

The equation of chord to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ bisected at the point $(x_{1}, y_{1})$ is given by

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

$∴ \frac{x \cdot 1}{4} + \frac{y \cdot 1 / 2}{2} - 1 = \frac{1}{4} + \frac{{(\frac{1}{2})}^{2}}{2} - 1$

$\Rightarrow x + y = 3 / 2$ ... (ii)

On solving equation (i) and (ii), we get

$x = \frac{6 \pm \sqrt{30}}{6} and y = \frac{3}{2} - (\frac{6 \pm \sqrt{30}}{6})$

Let $x_{1} = \frac{6 + \sqrt{30}}{6}, x_{2} = \frac{6 - \sqrt{30}}{6}$

and $y_{1} = \frac{3}{2} - (\frac{6 + \sqrt{30}}{6}), y_{2} = \frac{3}{2} - (\frac{6 - \sqrt{30}}{6})$

$∴$ The length of chord

$= \sqrt{(\frac{6 + \sqrt{30}}{6} - {(\frac{6 - \sqrt{30}}{6}))}^{2} + (\frac{3}{2} - (\frac{6 + \sqrt{30}}{6}) - \frac{3}{2} + {(\frac{6 - \sqrt{30}}{6}))}^{2}}$

$= \sqrt{\frac{30}{9} + \frac{30}{9}} = \sqrt{\frac{60}{9}}$

$= \frac{2}{3} \sqrt{15}$

Solution

Q.
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: [2025]

1 $\frac{1}{3} \sqrt{15}$

2 $\sqrt{15}$

3 $\frac{2}{3} \sqrt{15}$

4 $\frac{5}{3} \sqrt{15}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The length of the chord of the ellipse x24+y22=1, whose mid-point is (1,12), is: [2025]

1 1315

2 15

3 2315