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  • If the pair of lines joining origin to the points of intersection of    <img alt="\mathrm {\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

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

\mathrm {(a l, b m)\: \: lies \: \: on \: \: x^2=(n+y)(n-y)}\\


Option: 2

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


Option: 3

\mathrm {The \: \: product \: \: of \: \: a l+i b m \: \: and \: \: its \: \: conjugate \: \: is \: \: n^2}\\


Option: 4

\mathrm {all \: \: of \: \: these}


Answers (1)

best_answer

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.

 

Posted by

Riya

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