JEE Main 2025 Jan 29 Shift 2, Mathematics Q9: Ellipse: Chord, Tangent and Normal

JEE Main2025Jan 29, Shift 2Mathematics

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ , whose mid point is $(\frac{5}{2}, \frac{1}{2})$ , then $α + β$ is equal to

A
$37$
B
$46$
C
$58$
D
$72$

Solution

For a chord of $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with midpoint $(x_{1}, y_{1})$ , the equation is $T = S_{1}$ :

$\frac{x x _{1}}{a ^{2}} + \frac{y y _{1}}{b ^{2}} = \frac{x _{1}^{2}}{a ^{2}} + \frac{y _{1}^{2}}{b ^{2}}$ .

With $(x_{1}, y_{1}) = (\frac{5}{2}, \frac{1}{2})$ , $a^{2} = 9$ , $b^{2} = 4$ :

$\frac{5 x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16} = \frac{100 + 9}{144} = \frac{109}{144}$ .

Multiplying through by $144$ gives $40 x + 18 y = 109$ . So $α = 40$ , $β = 18$ , and $α + β = 58$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q9: Ellipse: Chord, Tangent and Normal

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