If the mid point of a chord of the ellipse $$x216+y225=1$$ is (0$$,$$3) , then length of the chord is

Equation of the chord whose mid point is (0$$,$$3) is $$3y25$$−$$1=925$$−$$1$$ i.e. $$y=3$$ It intersects the ellipse $$x216+y225=1$$ at $$x216=1$$−$$925=1625$$⇒$$x=$$±$$165$$∴ length of the chord $$=325$$

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$If the mid point of a chord of the ellipse \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 is (0, 3), then length of the chord is$

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Solution

$Equation of the chord whose mid point is (0, 3) is$
$\begin{array}{l} \frac{3 y}{25} - 1 = \frac{9}{25} - 1 i.e. y = 3 \\ It intersects the ellipse \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 at \end{array}$
$\begin{array}{l} \frac{x^{2}}{16} = 1 - \frac{9}{25} = \frac{16}{25} \\ \Rightarrow x = \pm \frac{16}{5} \\ ∴ length of the chord = \frac{32}{5} \end{array}$

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Ellipse

Remember concepts with our Masterclasses.

If the mid point of a chord of the ellipse x216+y225=1 is (0,3) , then length of the chord is

Equation of the chord whose mid point is (0,3) is 3y25−1=925−1 i.e. y=3 It intersects the ellipse x216+y225=1 at x216=1−925=1625⇒x=±165∴ length of the chord =325

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$If the mid point of a chord of the ellipse \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 is (0, 3), then length of the chord is$