If the line x-2y=12 is tangent to the ellipse x^2a^2+ y^2b^2=1 at the point (3 , -92 ) then the length of the latus rectum of the ellipse is:

Question

Accepted Answer

Correct answer: A. (a) JEE Main 2019 Equation of tangent to x^2a^2+ y^2b^2=1 at (3 ,- 92) is, 3xa^2- 9y2b^2=1 But given equation of tangent is, x-2y=12 ∴ 3a^2= -92b^2 (-2 )= 112 (On comparing) ⇒a^2=3 12 and b^2= 9 124 ⇒a=6 and b=3 3 Therefore, latus rectum = 2b^2a= 2 276=9

$If the line x - 2 y = 12 is tangent to the ellipse \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 at the point (3, \frac{- 9}{2}) then the length of the latus rectum of the ellipse is:$

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