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  • The locus of the mid point of the chords of  <img alt="\mathrm{\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E%7B2%7

The locus of the mid point of the chords of  \mathrm{\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1}, which pass through a fixed point \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) is

Option: 1

A circle


Option: 2

An ellipse


Option: 3

A hyperbola


Option: 4

None of these


Answers (1)

Let \mathrm{Q}(\mathrm{h}, \mathrm{k}) be the mid-point of drawn chord. Equation o this chord will be \mathrm{\mathrm{T}=\mathrm{S}_{1}} i.e.

\mathrm{\frac{x h}{a^{2}}-\frac{y k}{b^{2}}=\frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}}. If it's passes through \mathrm{P\left(x_{1}, y_{1}\right)\: then \: =\frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}-\frac{x_{1} h}{a^{2}}+\frac{y_{1} k}{b^{2}}=0}.

Thus, locus is \mathrm{\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=0}

which is clearly a hyperbola.

Posted by

Ramraj Saini

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