The locus of the middle points of the chords | vedclass

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(A) Let $P(x_1, y_1)$ be the middle point of the chord of the hyperbola $3x^2 - 2y^2 + 4x - 6y = 0$.The equation of the chord with midpoint $(x_1, y_1

Vedclass · Accepted Answer

$3x - 4y = 4$

Vedclass · Answer

(A) Let $P(x_1, y_1)$ be the middle point of the chord of the hyperbola $3x^2 - 2y^2 + 4x - 6y = 0$.The equation of the chord with midpoint $(x_1, y_1)$ is given by $T = S_1$.$3xx_1 - 2yy_1 + 2(x + x_1) - 3(y + y_1) = 3x_1^2 - 2y_1^2 + 4x_1 - 6y_1$.Rearranging the terms to find the slope of the chord:$(3x_1 + 2)x - (2y_1 + 3)y + (2x_1 - 3y_1 - 3x_1^2 + 2y_1^2) = 0$.The slope of this chord is $m = \frac{3x_1 + 2}{2y_1 + 3}$.Since the chord is parallel to $y = 2x$,its slope must be $2$.$\frac{3x_1 + 2}{2y_1 + 3} = 2$$3x_1 + 2 = 4y_1 + 6$$3x_1 - 4y_1 = 4$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $3x - 4y = 4$.

The locus of the middle points of the chords of the hyperbola $3x^2 - 2y^2 + 4x - 6y = 0$ parallel to $y = 2x$ is

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