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  • The locus of mid-points of a focal chord of the ellipse <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%5E2%7

The locus of mid-points of a focal chord of the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} is

Option: 1

\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{e x}{a}}


Option: 2

\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{e x}{a}}


Option: 3

\mathrm{x^2+y^2=a^2+b^2}


Option: 4

None of these


Answers (1)

best_answer

Let\left ( h,k \right ) be the mid point of a focal chord. Then its equation is \mathrm{S_1=T \text { or } \frac{h x}{a^2}+\frac{k y}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}} 
This passes through \mathrm{(a e, 0), \therefore \frac{h a e}{a^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2} \text {. So, locus of }(h, k) \text { is } \frac{x e}{a}=\frac{x^2}{a^2}+\frac{y^2}{b^2}}
 

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