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QJEEM25Ma9201: JEE Main 2025 Mathematics - Class 11 JEEM25Ma9201 Solution: If is the equation of the chord of the ellipse , whose mid…

If αx+βy=109\alpha x+\beta y=109 is the equation of the chord of the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 , whose mid point is (52,12)\left(\frac{5}{2}, \frac{1}{2}\right) . then α+β\alpha+\beta is equal to :

Multiple Choice Options

  • A 37
  • B 46
  • C 72
  • D 58

Step-by-Step Verified Solution

JEE Main 2025 (Online) 29th January Evening Shift Mathematics - Ellipse Question 7 English Explanation

amp; Equation of chord T=S1amp;amp;52(x9)+12(y4)=2536+116amp;⇒5x18+y8=100+9144=109144amp;⇒40x+18y=109amp;⇒α=40,β=18amp;⇒α+β=58\begin{aligned}&\text { Equation of chord } \mathrm{T}=\mathrm{S}_1\\&\begin{aligned}& \frac{5}{2}\left(\frac{\mathrm{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}\end{aligned}

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