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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Ellipse

If α x+β y=109 is the equation of the chord of the ellipse...

If αx+βy=109istheequationofthechordoftheellipse\alpha x+\beta y=109 is the equation of the chord of the ellipse\frac{x^2}{9}+\frac{y^2}{4}=1,whosemidpointis(52,12), whose mid point is \left(\frac{5}{2}, \frac{1}{2}\right), then α+β\alpha+\beta is equal to :If α x+β y=109 is the equation of the chord of the ellipse (x^2)/(9)+(y^2)/(4)=1 , whose mid point is ((5)/(2),

Options

  1. A58
  2. B46
  3. C37
  4. D72

Correct answer

A. 58

Step-by-step solution

If α x+β y=109 is the equation of the chord of the ellipse (x^2)/(9)+(y^2)/(4)=1 , whose mid point is ((5)/(2),
Equation of chord T =S1=\mathrm{S}_152(x9)+12(y4)=2536+116⇒5x18+y8=100+9144=109144⇒40x+18y=109⇒α=40,β=18⇒α+β=58\begin{aligned} & \frac{5}{2}\left(\frac{x}{9}\right)+\frac{1}{2}\left(\frac{\mathrm{y}}{4}\right)=\frac{25}{36}+\frac{1}{16} \\ & \Rightarrow \frac{5 \mathrm{x}}{18}+\frac{\mathrm{y}}{8}=\frac{100+9}{144}=\frac{109}{144} \\ & \Rightarrow 40 \mathrm{x}+18 \mathrm{y}=109 \\ & \Rightarrow \alpha=40, \beta=18 \\ & \Rightarrow \alpha+\beta=58 \end{aligned}$$
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