Find the equation of the chord of the ellipse | vedclass

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(A) The equation of a chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ bisected at $(x_1, y_1)$ is given by $T = S_1$.Here,the equation of

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$4x + 5y - 13 = 0$

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(A) The equation of a chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ bisected at $(x_1, y_1)$ is given by $T = S_1$.Here,the equation of the ellipse is $2x^2 + 5y^2 = 20$,which can be written as $\frac{x^2}{10} + \frac{y^2}{4} = 1$.Given point $(x_1, y_1) = (2, 1)$.$T = 2x(x_1) + 5y(y_1) - 20 = 2x(2) + 5y(1) - 20 = 4x + 5y - 20$.$S_1 = 2(x_1)^2 + 5(y_1)^2 - 20 = 2(2)^2 + 5(1)^2 - 20 = 2(4) + 5(1) - 20 = 8 + 5 - 20 = -7$.Equating $T = S_1$,we get $4x + 5y - 20 = -7$.$4x + 5y - 20 + 7 = 0$.$4x + 5y - 13 = 0$.

Find the equation of the chord of the ellipse $2x^2 + 5y^2 = 20$ which is bisected at the point $(2, 1)$.

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