The differential equation ydx - y 2 dy - xdy and y 1-1 represents a parabola whoseA- Vertex is 0-0B- axis is x-1C- length of latus rectum is 4D- tangent at the vertex is x-1

The correct option is D tangent at the vertex is x = 1 Given equation ydx xdy/y^2 + dy = 0 ⇒ x/y + y = c, but y(1) = 1 ⇒ c = 2 ∴ x/y + y = 2 ⇒ x + y^2 ...

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Standard XII

Mathematics

Standard Equation of Parabola

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Question

The differential equation ydx + $y^{2}$ dy = xdy and y(1) = 1 represents a parabola whose

Vertex is (0, 0)

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axis is x = 1

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length of latus rectum is 4

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tangent at the vertex is x = 1

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Solution

The correct option is D
tangent at the vertex is x = 1

Given equation

$\frac{y d x - x d y}{y^{2}}$ + dy = 0

$\Rightarrow$ $\frac{x}{y}$ + y = c,

but y(1) = 1 $\Rightarrow$ c = 2

$∴$ $\frac{x}{y}$ + y = 2

$\Rightarrow$ x + $y^{2}$ = 2y

$\Rightarrow$ ${(y - 1)}^{2}$ = - (x-1)

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The differential equation ydx +y2dy = xdy and y(1) = 1 represents a parabola whose

The correct option is D tangent at the vertex is x = 1 Given equation ydx−xdyy2 + dy = 0 ⇒ xy + y = c, but y(1) = 1 ⇒ c = 2 ∴ xy + y = 2 ⇒ x + y2 = 2y ⇒ (y−1)2 = - (x-1)

The differential equation ydx + $y^{2}$ dy = xdy and y(1) = 1 represents a parabola whose

The correct option is D
tangent at the vertex is x = 1

Given equation

$\frac{y d x - x d y}{y^{2}}$ + dy = 0

$\Rightarrow$ $\frac{x}{y}$ + y = c,

but y(1) = 1 $\Rightarrow$ c = 2

$∴$ $\frac{x}{y}$ + y = 2

$\Rightarrow$ x + $y^{2}$ = 2y

$\Rightarrow$ ${(y - 1)}^{2}$ = - (x-1)