#### If the pair of lines joining origin to the points of intersection of    $\mathrm {\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$   by the line   $\mathrm {l x+m y=n}$  be coincident then  Option: 1 Option: 2 Option: 3 Option: 4

From given line, we have   $\mathrm {\frac{l x+m y}{n}=1}$ ....................(1)

By making the given ellipse homogeneous with (1), we get

\begin{aligned} & \mathrm {\frac{x^2}{a^2}+\frac{y^2}{b^2}=\left(\frac{l x+m y}{n}\right)^2} \\ & \mathrm {\Rightarrow\left(\frac{1}{a^2}-\frac{l^2}{n^2}\right) x^2-\frac{2 l m}{n^2} \cdot x y+\left(\frac{1}{b^2}-\frac{m^2}{n^2}\right) y^2=0} \end{aligned}......................(2)

Now, (2) will represent a pair of coincident lines if

$\mathrm {\left(-\frac{l m}{n^2}\right)^2=\left(\frac{1}{a^2}-\frac{l^2}{n^2}\right)\left(\frac{1}{b^2}-\frac{m^2}{n^2}\right)}$

[From $\mathrm {h^2=a b}$ for coincident lines]

\begin{aligned} & \mathrm {\Rightarrow \frac{l^2 m^2}{n^4}=\frac{n^2-a^2 l^2}{a^2 n^2} \cdot \frac{n^2-b^2 m^2}{b^2 n^2} }\\\\ & \mathrm{ \Rightarrow a^2 b^2 l^2 m^2=n^4-b^2 m^2 n^2-a^2 l^2 n^2+a^2 b^2 l^2 m^2 }\\\\ &\mathrm { \Rightarrow a^2 l^2+b^2 m^2=n^2} \end{aligned}............(3)

$\mathrm {\Rightarrow a^2 l^2=(n+b m)(n-b m)}$  shows that  $\mathrm {(a l, b m) }$  lies on

$\mathrm {x^2=(n+y)(n-y) \Rightarrow( a) }$  is correct

From (3),     $\mathrm { \frac{\left(-a^2 l\right)^2}{a^2}+\frac{\left(-b^2 m\right)^2}{b^2}=n^2 }$ shows that

$\mathrm { \left(-a^2 l,-b^2 m\right) }$  lies on    $\mathrm {\frac{x^2}{a^2}+\frac{y^2}{b^2}=n^2 \Rightarrow(\mathrm{b})}$  is correct

Further, product of   $\mathrm {(a l+i b m)}$  and its  conjugate

$\mathrm {=(a l+i b m)(a l-i b m)=a^2 l^2+b^2 m^2=n^2}$  [from (3)]

$\mathrm {\Rightarrow(\mathrm{3})}$  is correct.