Question

The equation of the line, passing through the centre and bisecting the chord $7x+y-1=0$ of the ellipse $\frac{x^2}{1}+\frac{y^2}{7}=1,$ is

• A. $x-7y=0$
• B. $7x-y=0$
• C. $x-y=0$
• D. $x+y=0$

Answer 1

The correct option is C $x-y=0$


Given, equation of the ellipse is
$\frac{x^2}{1}+\frac{y^2}{7}=1$
Equation of chord $QR$ with mid point at $(h,k)$ is
$T=S_1$
$\Rightarrow \frac{hx}{1}+\frac{ky}{7}=\frac{h^2}{1}+\frac{k^2}{7}\dots \left(1\right)$
But given, equation of $QR$ is
$7x+y=1\dots \left(2\right)$
$\because$ Line $\left(1\right)$ and $\left(2\right)$ are identical
$\therefore \frac{h}{7}=\frac{k}{7}\Rightarrow h=k$
Hence, equation of the line joining $(0,0)$ and $(h,k)$ is $y-x=0.$