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Find the equation of the line passing through the centre and which bisects the chord $\mathrm{7 x+y-20=0}$ of the ellipse $\mathrm{x^2+\frac{y^2}{7}=40.}$
Option: 1
$\mathrm{y=2x}$

Option: 2
$\mathrm{y=-2x}$

Option: 3
$\mathrm{y=x}$

Option: 4
$\mathrm{y=-x}$

Answers (1)

The centre of the ellipse is (0,0).

Let the line be $\mathrm{y=m x.}$

The ends $\mathrm{\left(x_1, y_1\right)}$ and $\mathrm{\left(x_2, y_2\right)}$ of the chords are found by solving $\mathrm{7 x+y-20=0}$ and

$\mathrm{x^2+\frac{y^2}{7}=40}$

We get, $\mathrm{x^2+\frac{1}{7}(20-7 x)^2=40 \Rightarrow 56 x^2-280 x+120=0}$

Its roots are $\mathrm{x_1}$ and $\mathrm{x_2}$

So, $\mathrm{x_1+x_2=\frac{280}{56}=5}$

Also, $\mathrm{y=20-7 x}$ or, $\mathrm{y_1+y_2=40-7\left(x_1+x_2\right)=5}$

The middle point of the chord is $\mathrm{\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)}$ i.e. $\mathrm{\left(\frac{5}{2}, \frac{5}{2}\right)}$

As this lies on $\mathrm{y=m x, m=1}$

So, the equation of the line is $\mathrm{y=x.}$

Posted by

Ramraj Saini

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