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  • What will be the equation of that chord of ellipse <img alt="\mathrm{\frac{x^2}{36}+\frac{y^2}{9}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B36%7

What will be the equation of that chord of ellipse \mathrm{\frac{x^2}{36}+\frac{y^2}{9}=1} which passes from the point  \mathrm{\left ( 2,1 \right )}
and bisected on the point 

Option: 1

\mathrm{x+y=2}


Option: 2

\mathrm{x+y=3}


Option: 3

\mathrm{x+2 y=1}


Option: 4

\mathrm{x+2 y=4}


Answers (1)

best_answer

Let required chord meets to ellipse on the points P and Q whose coordinates are \mathrm{\left ( x,y \right )} and \mathrm{\left(x_2, y_2\right)} 
 

respectively

    Point (2,1) is mid point of chord PQ
\mathrm{\therefore \quad 2=\frac{1}{2}\left(x_1+x_2\right) \text { or } x_1+x_2=4 \text { and } 1=\frac{1}{2}\left(y_1+y_2\right) \text { or } y_1+y_2=2}
Again points \mathrm{\left(x_1, y_1\right) \text { and }\left(x_2, y_2\right)} are situated on ellipse; \mathrm{\therefore \frac{x_1^2}{36}+\frac{y_1^2}{9}=1 \text { and } \frac{x_2^2}{36}+\frac{y_2^2}{9}=1}
On subtracting \mathrm{\frac{x_2^2-x_1^2}{36}+\frac{y_2^2-y_1^2}{9}=0 \quad \frac{y_2-y_1}{x_2-x_1}=-\frac{\left(x_2+x_1\right)}{4\left(y_2+y_1\right)}=\frac{-4}{4 \times 2}=\frac{-1}{2}}
∴  Gradient of chord \mathrm{P Q=\frac{y_2-y_1}{x_2-x_1}=\frac{-1}{2}}
Therefore, required equation of chord PQ is as follows, \mathrm{y-1=-\frac{1}{2}(x-2) \text { or } x+2 y=4}
Alternative:S_1=T     (If mid point of chord is known)
\mathrm{\therefore \frac{2^2}{36}+\frac{1^2}{9}-1=\frac{2 x}{36}+\frac{1 y}{9}-1 \Rightarrow x+2 y=4}

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