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  • The equation of the line passing through the centre and bisecting the chord 7 x+y-1=0 of the ellipse <img alt="\mathrm{x^2+\frac{y^2}{7}=1}" src="https://entrancecorner.oncodecogs.com/gif.late

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

Option: 1

x=0


Option: 2

y=0


Option: 3

y+x=0


Option: 4

y-x=0


Answers (1)

best_answer

Let (h, k) be the mid-point of the chord 7x + y – 1 = 0     ...(i)

\mathrm{\text { Equation of the chord of the ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { bisected }} at (h, k) is

\mathrm{\frac{h x}{1}+\frac{k y}{7}=\frac{h^2}{1}+\frac{k^2}{7}}             ....[ii]

Now, equations (i) and (ii) represent the same straight line.

\mathrm{\therefore \quad \frac{h}{7}=\frac{k}{7} \Rightarrow h=k}

Equation of the line joining (0, 0) and (h, k) is y – x = 0

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