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  • The mid-point of the chord  of the ellipse <img alt="\mathrm{\frac{x^2}{4

The mid-point of the chord \mathrm{9 x-4 y=13} of the ellipse \mathrm{\frac{x^2}{4}+\frac{y^2}{9}=1} is (a, b) then value of a-b is--

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

Let (h, k) is mid point of chord then equation of chord is

\mathrm{\begin{aligned} & \Rightarrow \frac{h x}{4}+\frac{k y}{9}=\frac{h^2}{4}+\frac{k^2}{9} \\ & \Rightarrow \frac{9 h x+4 k y}{36}=\frac{9 h^2+4 k^2}{36} \\ & \Rightarrow 9 h x+4 k y=9 h^2+4 k^2 \end{aligned}}                ,,,,[i]

Compare equation (i) with 9x – 4y = 13,
9h = 9, 4k = –4
⇒ h = 1 ⇒ k = –1
The mid point is (1, –1).
So, a = 1, b = –1
The value of a – b = 1 + 1 = 2

 

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vinayak

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