Consider the curve f x - y -0 which satisfies the differential equation dy - dx -1- x - y 2-4-0 such that y 1-1- If f x - y represents a conic- then the length of its latus rectum is

Dydx+1x y^2+4=0 ⇒dxdy+x=y^2 4 ⇒ xe^y= ∫ e^y(y^2 4)dy ⇒ xe^y=e^y(y^2 4) e^y(2y)+e^y· 2+c ⇒ x=(y^2 4 2y+2)+ce^ y ⇒ x=(y 1)^2 3+ce^ y Curve passes through (1, 1). ...

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Standard XII

Mathematics

Integrating Factor

Consider the ...

Question

Consider the curve $f (x, y) = 0$ which satisfies the differential equation $\frac{d y}{d x} + \frac{1}{x - y^{2} + 4} = 0$ such that $y (1) = - 1$ . If $f (x, y)$ represents a conic, then the length of its latus rectum is

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Solution

$\frac{d y}{d x} + \frac{1}{x - y^{2} + 4} = 0$
$\Rightarrow \frac{d x}{d y} + x = y^{2} - 4 \Rightarrow x e^{y} = \int e^{y} (y^{2} - 4) d y \Rightarrow x e^{y} = e^{y} (y^{2} - 4) - e^{y} (2 y) + e^{y} \cdot 2 + c \Rightarrow x = (y^{2} - 4 - 2 y + 2) + c e^{- y} \Rightarrow x = (y - 1)^{2} - 3 + c e^{- y}$
Curve passes through $(1, - 1) .$
$\Rightarrow 1 = 4 - 3 + c e^{1}$
$\Rightarrow c = 0$
$∴ (y - 1)^{2} = (x + 3)$

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Consider the curve f(x,y)=0 which satisfies the differential equation dydx+1x−y2+4=0 such that y(1)=−1. If f(x,y) represents a conic, then the length of its latus rectum is

dydx+1x−y2+4=0 ⇒dxdy+x=y2−4⇒xey=∫ey(y2−4)dy⇒xey=ey(y2−4)−ey(2y)+ey⋅2+c⇒x=(y2−4−2y+2)+ce−y⇒x=(y−1)2−3+ce−y Curve passes through (1,−1). ⇒1=4−3+ce1 ⇒c=0 ∴(y−1)2=(x+3)

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