The mid-point of a chord of the ellipse x^2+4 | vedclass

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(A) The equation of a chord of a conic $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.Given equation of ellipse: $S = x^2+4y^2-2x+20y = 0$.Mid-

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$x-6y=26$

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(A) The equation of a chord of a conic $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.Given equation of ellipse: $S = x^2+4y^2-2x+20y = 0$.Mid-point $(x_1, y_1) = (2, -4)$.$T = x(x_1) + 4y(y_1) - (x+x_1) + 10(y+y_1) = 0$.Substituting $(x_1, y_1) = (2, -4)$:$T = 2x + 4y(-4) - (x+2) + 10(y-4) = 2x - 16y - x - 2 + 10y - 40 = x - 6y - 42$.$S_1 = (2)^2 + 4(-4)^2 - 2(2) + 20(-4) = 4 + 64 - 4 - 80 = -16$.Equating $T = S_1$:$x - 6y - 42 = -16$.$x - 6y = 42 - 16$.$x - 6y = 26$.

The mid-point of a chord of the ellipse $x^2+4y^2-2x+20y=0$ is $(2,-4)$. The equation of the chord is

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