The length of the chord of the ellipse , whose mid point is , is equal to : [2024]

Question

The length of the chord of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid point is $(1, \frac{2}{5})$ , is equal to : [2024]

Smart Achievers · Accepted Answer

(2)

Ellipse : $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ ... (i)

Equation of chord having mid point $(x_{1}, y_{1})$ is

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

$\Rightarrow \frac{x}{25} + \frac{y}{16} (\frac{2}{5}) = \frac{1}{25} + \frac{1}{16} (\frac{4}{25}) \Rightarrow \frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{1}{100}$

$\Rightarrow 8 x + 5 y = 10$

$\Rightarrow 8 x = 5 (2 - y)$ ...(ii)

From equation (i) & (ii), we get

$x = 1 \pm \frac{\sqrt{19}}{2}, y = \frac{2}{5} (1 \mp 2 \sqrt{19)}$

Length of Chord

$= \sqrt{{[1 + \frac{\sqrt{19}}{2} - (1 - \frac{\sqrt{19}}{2})]}^{2} + {[\frac{2}{5} (1 - 2 \sqrt{19}) - \frac{2}{5} (1 + 2 \sqrt{19})]}^{2}}$

$= \sqrt{19 + \frac{64}{25} \times 19} = \frac{\sqrt{1691}}{5}$

Solution

Q.
The length of the chord of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ , whose mid point is $(1, \frac{2}{5})$ , is equal to : [2024]

1 $\frac{\sqrt{1541}}{5}$

2 $\frac{\sqrt{1691}}{5}$

3 $\frac{\sqrt{1741}}{5}$

4 $\frac{\sqrt{2009}}{5}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The length of the chord of the ellipse x225+y216=1, whose mid point is (1, 25), is equal to : [2024]

1 15415

2 16915

3 17415