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10-2. Parabola, Ellipse, Hyperbola
hard
If $\alpha x+\beta y=109$ is the equation of the chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, whose mid point is $\left(\frac{5}{2}, \frac{1}{2}\right)$, then $\alpha+\beta$ is equal to
A$37$
B$46$
C$58$
D$72$
(JEE MAIN-2025)
Solution

Equation of chord $T = S _1$
$\frac{5}{2}\left(\frac{x}{9}\right)+\frac{1}{2}\left(\frac{y}{4}\right)=\frac{25}{36}+\frac{1}{16}$
$\Rightarrow \frac{5 x}{18}+\frac{y}{8}=\frac{100+9}{144}=\frac{109}{144}$
$\Rightarrow 40 x+18 y=109$
$\Rightarrow \alpha=40, \beta=18$
$\Rightarrow \alpha+\beta=58$
Standard 11
Mathematics