Latus rectum of conic satisfying differential equation, x dy + y dx = 0 and passing through p

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10-2. Parabola, Ellipse, Hyperbola

normal

Latus rectum of the conic satisfying the differential equation, $ x dy + y dx = 0$ and passing through the point $ (2, 8) $ is :

$4 \sqrt 2 $

$8$

$8 \sqrt 2 $

$16$

Solution

$\frac{{dy}}{y}\,\, + \,\frac{{dx}}{x}\, = \,0 $ $\Rightarrow$ $ ln$ $ xy = c$ $\Rightarrow$ $ xy = c$
passes through $(2,8)$ $ \Rightarrow$ $ c = 16$
$xy =16$ $LR = 2a(e^2 – 1) = 2a$
solving with $y = x$
vertex is $(4, 4)$
distance from centre to vertex = $4\sqrt 2 \,$
$L.R. = $ length of $TA =8\sqrt 2 \,$

Standard 11

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Latus rectum of the conic satisfying the differential equation, $ x dy + y dx = 0$ and passing through the point $ (2, 8) $ is :

Solution

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