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If   \mathrm { y=m x}  and  \mathrm { y=m^{\prime} x} are two diameters of the ellipse   \mathrm { \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}   such that each bisects the chord parallel to the other, then  \mathrm {m m^{\prime}=} 
 

Option: 1

\mathrm { \frac{b^2}{a^2}}


Option: 2

\mathrm { \frac{-2 b^2}{a^2}}


Option: 3

\mathrm { \frac{-b^2}{a^2}}


Option: 4

\mathrm { \frac{-b^2}{2 a^2}}


Answers (1)

best_answer

Let   \mathrm {\left(x^{\prime}, y^{\prime}\right)}  be the midpoint of a chord parallel to  \mathrm {y=m x,}  then equation of this chord is   \mathrm {\frac{x x^{\prime}}{a^2}+\frac{y y^{\prime}}{b^2}=\frac{x^{\prime}}{a^2}+\frac{y^{\prime 2}}{b^2} }    so that
\mathrm { m=-\frac{b^2 x^{\prime}}{a^2 y^{\prime}} } 
Also, \mathrm { \left(x^{\prime}, y^{\prime}\right) lies\: \: on\: \: y=m^{\prime} x \Rightarrow y^{\prime}=m^{\prime} x^{\prime} }

From Eqs. (i) and (ii), we get

\mathrm { m m^{\prime}=-\frac{b^2 x^{\prime}}{a^2 y^{\prime}} \times \frac{y^{\prime}}{x^{\prime}} \quad \text { or } \quad m m^{\prime}=-\frac{b^2}{a^2} }

Posted by

Ritika Kankaria

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