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

3 x - 4 y = 4

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4 x - 4 y = 3

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3 y + 4 x + 4 = 0

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3 x + 4 y = 2

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Solution

The correct option is B $3 x - 4 y = 4$
Let $(x_{1}, y_{1})$ be the mid point of a chord.
Equation of chord is given by $T = S_{1}$
i.e. $3 x x_{1} - 2 y y_{1} + 2 (x + x_{1}) - 3 (y + y_{1}) = 3 x x_{1} - 2 y y_{1} + 4 x_{1} - 6 y_{1}$
Slope of this chord = $2 = \frac{2 + 3 x_{1}}{3 + 2 y_{1}}$
So, $2 + 3 x_{1} = 6 + 4 y_{1}$
In other words, the locus of $(x_{1}, y_{1})$ is $3 x - 4 y = 4$

Suggest Corrections

Similar questions

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

y = 2 x

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

y = 2 x

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

y = 2 x

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

y = 2 x

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