- Home
- Standard 11
- Mathematics
The locus of the poles of normal chords of an ellipse is given by
$\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {({a^2} - {b^2})^2}$
$\frac{{{a^3}}}{{{x^2}}} + \frac{{{b^3}}}{{{y^2}}} = {({a^2} - {b^2})^2}$
$\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {({a^2} + {b^2})^2}$
$\frac{{{a^3}}}{{{x^2}}} + \frac{{{b^3}}}{{{y^2}}} = {({a^2} + {b^2})^2}$
Solution
(a) Let the equation of the ellipse is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$…..$(i)$
Let $(h,\,k)$ be the poles.
Now polar of $(h,\,k)$ w.r.t. the ellipse is given by $\frac{{xh}}{{{a^2}}} + \frac{{yk}}{{{b^2}}} = 1$ …..$(ii)$
If it is a normal to the ellipse then it must be identical with $ax\,\sec \theta – \,by\,{\rm{cosec}}\,\theta = {a^{\rm{2}}} – {b^2}$…..$(iii)$
Hence comparing $(ii)$ and $(iii),$ we get
$\frac{{(h/{a^2})}}{{a\,\sec \theta }} = \frac{{(k/{b^2})}}{{ – b\,\cos ec\theta }} = \frac{1}{{({a^2} – {b^2})}}$
==> $\cos \theta = \frac{{{a^3}}}{{h\,({a^2} – {b^2})}}$ and $\sin \theta = \frac{{{b^3}}}{{k({a^2} – {b^2})}}$
Squaring and adding we get,
$1 = \frac{1}{{{{({a^2} – {b^2})}^2}}}\left( {\frac{{{a^6}}}{{{h^2}}} + \frac{{{b^6}}}{{{k^2}}}} \right)\,$
Required locus of $(h,\,k)$ is $\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {({a^2} – {b^2})^2}.$
