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  • The locus of the middle points of chords of hyperbola <img alt="\mathrm{3 x^{2}-2 y^{2}+4 x-6 y=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3%20x%5E%7B2%7D-2%20y%5E%7B2%7D&

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

Option: 1

\mathrm{3 x-4 y=4}


Option: 2

\mathrm{3 y-4 x+4=0}


Option: 3

\mathrm{4 x-4 y=3}


Option: 4

\mathrm{3 x-4 y=2}


Answers (1)

best_answer

Let \mathrm{(h, k)} be the midpoint of a chord of the hyperbola \mathrm{3 x^{2}-2 y^{2}+4 x-6 y=0}.

Then the equation of the chord is
\mathrm{3 \mathrm{hx}-2 \mathrm{ky}+2(\mathrm{x}+\mathrm{h})-3(\mathrm{y}+\mathrm{k})=3 \mathrm{~h}^{2}-2 \mathrm{k}^{2}+4 \mathrm{~h}-6 \mathrm{k} \quad [Using \left.\mathrm{T}=\mathrm{S}^{\prime}\right]}

\mathrm{\Rightarrow(3 \mathrm{~h}+2) \mathrm{x}-(2 \mathrm{k}+3) \mathrm{y}-3 \mathrm{~h}^{2}+2 \mathrm{k}^{2}-2 \mathrm{~h}+3 \mathrm{k}=0}

This is parallel to \mathrm{\mathrm{y}=2 \mathrm{x}}, therefore

\mathrm{\frac{-(3 \mathrm{~h}+2)}{-(2 \mathrm{k}+3)}=2 \Rightarrow 3 \mathrm{~h}+2=4 \mathrm{k}+6 \Rightarrow 3 \mathrm{~h}-4 \mathrm{k}=4}

Hence, locus of \mathrm{(h, k)} is \mathrm{3 x-4 y=4}.

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