Let M (1/2, 1) be the mid-point of a chord to the Ellipse x^2/2+y^2/4, then the length of chord is

Question

Let \(M\left(\frac{1}{2}, 1\right)\) be the mid-point of a chord to the Ellipse \(\frac{x^2}{2}+\frac{y^2}{4}=1,\) then the length of chord is

(1) \(\frac{2}{3} \sqrt{5}\)

(2) \(\frac{\sqrt{5}}{3}\)

(3) \(2 \sqrt{\frac{5}{3}}\)

(4) \(\frac{\sqrt{5}}{2}\)

Answer 1

Correct option is (3) \(2 \sqrt{\frac{5}{3}}\)   

\(\frac{x^2}{2}+\frac{y^2}{4}=1\)  

The equation of chord bisected at \((\frac{1}{2},1)\) will be

T = \(S_1.\)  

\(\Rightarrow \frac{x}{2}\left(\frac{1}{2}\right)+\frac{y}{4}(1)-1=\frac{\left(\frac{1}{2}\right)^2}{2}+\frac{1^2}{4}-1 \)   

\(\Rightarrow \frac{x}{4}+\frac{y}{4}=\frac{1}{8}+\frac{1}{4} \)  

\(\Rightarrow x+y=\frac{3}{2} \Rightarrow x_1+y_1=\frac{3}{2} \)