The line `l x+m y+n=0` is a normal to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` . then prove that `(a^2)/(l^2)+(b^2)/(m^2)=((a^2-b^2)^2)/(n^2)`

Options

A. n
B. `n^(2)`
C. `n^(3)`
D. None of these

Answer

B

Explanation

The equation of any normal to
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is
`axcosphi+bycotphi=a^(2)+b^(2)`
`impliesaxcosphi+bycotphi-(a^(2)+b^(2))=0`
the straight line `lx+my-n=0` will be normal to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. then Eq. (i) and `lx+my-n=0` represent the same line.
`therefore(acosphi)/(l)=(bcotphi)/(m)=(a^(2)+b^(2))/(n)`
`impliesecphi=(na)/(l(a^(2)+b^(2)))`
and `tanphi=(nb)/(m(a^(2)+b^(2)))`
`becausesec^(2)phi-tan^(2)phi=1`
`therefore(n^(2)a^(2))/(l^(2)(a^(2)+b^(2))^(2))-(n^(2)b^(2))/(m^(2)(a^(2)+b^(2))^(2))=1`
`implies(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(n^(2))`
But given equation of normal is
`(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(k)`
`thereforek=n^(2)`