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

Question

Latus rectum of the conic satisfying the differential equation, x d y+y d x=0 and passing through the point (2,8) is : (A) 4 √2 (B) 8 (C) 8 √2 (D) 16

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/25d0fdbf0ac72e054e4bd84b50679355-.png" /> ( dy / y )+( dx / x )=0 ⇒ ln xy = c ⇒ xy = c text passes through (2,8) ⇒ c =16 xy =16 LR =2 a ( e 2-1)=2 a ( e =√2) text solving with y = x text vertex is (4,4) text distance from centre to vertex =4 √2 L.R. = length of TA =8 √2

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

$42$

8

$82$

16

$\frac{d y}{y} + \frac{d x}{x} = 0 \Rightarrow ln x y = c \Rightarrow x y = c$
$passes through (2, 8) \Rightarrow c = 16$
$x y = 16 L R = 2 a (e^{2} - 1) = 2 a (e = 2)$
$solving with y = x$
$vertex is (4, 4)$
$distance from centre to vertex = 42$
L.R. $=$ length of $T A = 82$

Q. Latus rectum of the conic satisfying the differential equation, xdy+ydx=0 and passing through the point (2,8) is :

42​

8

82​

16

ydy​+xdx​=0⇒lnxy=c⇒xy=c passes through (2,8)⇒c=16 xy=16LR=2a(e2−1)=2a(e=2​) solving with y=x vertex is (4,4) distance from centre to vertex =42​ L.R. = length of TA=82​

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

$42$

$82$

$\frac{d y}{y} + \frac{d x}{x} = 0 \Rightarrow ln x y = c \Rightarrow x y = c$
$passes through (2, 8) \Rightarrow c = 16$
$x y = 16 L R = 2 a (e^{2} - 1) = 2 a (e = 2)$
$solving with y = x$
$vertex is (4, 4)$
$distance from centre to vertex = 42$
L.R. $=$ length of $T A = 82$