Question

The midpoint of a chord of the ellipse $x^2+4y^2-2x+20y=0$ is $(2,-4)$. The equation of the chord is

• A. $x-6y=26$
• B. $x+6y=26$
• C. $6x-y=26$
• D. $6x+y=26$

Answer 1

The correct option is A $x-6y=26$

$x^2+4y^2-2x+20y=0$......................(1)

Equation (1) can be reduced as

$\frac{(x-1{)}^2}{26}+\frac{(y+\frac{5}{2}{)}^2}{\frac{26}{4}}=1$

Equation of chord of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with midpoint $(x_1,y_1)$ is

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}$

So, here, equn. of chord is

$\frac{(x-1)(x_1-1)}{26}+\frac{(y+\frac{5}{2})(y_1+\frac{5}{2})}{\frac{26}{4}}=\frac{(x_1-1{)}^2}{26}+\frac{(y_1+\frac{5}{2}{)}^2}{\frac{26}{4}}$

$\frac{(x-1)}{26}+\frac{4\times \frac{-3}{2}(y+\frac{5}{2})}{26}=\frac{1}{26}+\frac{9}{26}$

$(x-1)-6(y+\frac{5}{2})=10$

$\Rightarrow x-1-6y-15=10$

$\Rightarrow x-6y=26$