Question

The line $lx+my=1$ meets the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in the points $P$ and $Q$, then the mid point of chord $PQ$ is, where $k=a^2{l}^2+b^2{m}^2$

• A. $\left(\frac{a^{2}l}{k},\frac{b^{2}m}{k}\right)$
• B. $\left(\frac{-a^{2}l}{k},\frac{-b^{2}m}{k}\right)$
• C. $\left(\frac{b^{2}l}{k},\frac{a^{2}m}{k}\right)$
• D. $\left(\frac{-b^{2}l}{k},\frac{-a^{2}m}{k}\right)$

Answer 1

Answer: (A)

$\left(\frac{a^{2}l}{k},\frac{b^{2}m}{k}\right)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Or
$b^2{x}^2+a^2{y}^2=a^2{b}^2$
Now the line $lx+my-1$ cuts the ellipse at two points.
$y=\frac{1-lx}{m}$
Substituting in the equation of ellipse gives us
$b^2{x}^2+a^2(\frac{1-lx}{m}{)}^2=a^2{b}^2$
$m^2{b}^2{x}^2+a^2(1-lx{)}^2=a^2{b}^2{m}^2$
$m^2{b}^2{x}^2+a^2[1+l^2{x}^2-2lx]=a^2{b}^2{m}^2$
$x^2(m^2{b}^2+a^2{l}^2)-2a^{2}lx+a^2-a^2{b}^2{m}^2=0$
$x=\frac{2a^{2}l\pm \sqrt{4a^4{l}^2-4(m^2{b}^2+a^2{l}^2)(a^2-a^2{b}^2{m}^2)}}{2(m^2{b}^2+a^2{l}^2)}$
$x=\frac{a^{2}l\pm \sqrt{a^4{l}^2-(m^2{b}^2+a^2{l}^2)(a^2-a^2{b}^2{m}^2)}}{(m^2{b}^2+a^2{l}^2)}$
$x=\frac{a^{2}l\pm \sqrt{a^4{l}^2-a^2(a^2{l}^2+m^2{b}^2)(1-m^2{b}^2)}}{(m^2{b}^2+a^2{l}^2)}$
Let the $x-coordinate$ of P and Q be $x_1,x_2$, then
$x_1+x_2=\frac{-B}{A}=\frac{2a^{2}l}{a^2{l}^2+m^2{b}^2}$
Hence
$\frac{x_1+x_2}{2}=\frac{a^{2}l}{a^2{l}^2+m^2{b}^2}$
$x^{\prime }=\frac{a^{2}l}{a^2{l}^2+m^2{b}^2}$
Thus $y^{\prime }=\frac{1-l{x}^{\prime }}{m}$
$=\frac{1-l\left(\frac{a^{2}l}{a^2{l}^2+m^2{b}^2}\right)}{m}$
$=\frac{a^2{l}^2+m^2{b}^2-a^2{l}^2}{m(a^2{l}^2+m^2{b}^2)}$
$=\frac{m^2{b}^2}{m(a^2{l}^2+m^2{b}^2)}$
$=\frac{b^{2}m}{a^2{l}^2+m^2{b}^2}$
Hence
$(x^{\prime },y^{\prime })=(\frac{a^{2}l}{a^2{l}^2+m^2{b}^2},\frac{b^{2}m}{a^2{l}^2+m^2{b}^2})$