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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Find the lengths of the principal axes. co-ordinates of the focii equations of directrices length of the latus rectum distance between focii distance between directrices of the ellipse: 3x2 + 4y2 = 1 - Mathematics and Statistics

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प्रश्न

Find the 

  1. lengths of the principal axes. 
  2. co-ordinates of the focii 
  3. equations of directrices 
  4. length of the latus rectum
  5. distance between focii 
  6. distance between directrices of the ellipse:

3x2 + 4y2 = 1

बेरीज

उत्तर

The equation of the ellipse is 3x2 + 4y2 = 1

i.e., `x^2/((1/3)) + y^2/((1/4))` = 1

Comparing with `x^2/"a"^2 + y^2/"b"^2` = 1, we get

a2 = `1/3`, b2 = `1/4`

∴ a = `1/sqrt3`, b = `1/2`

∴ a > b

i. Length of major axis = 2a = `2/sqrt(3)`

Length of minor axis = 2b = `2(1/2)` = 1

ii. Eccentricity = e = `sqrt("a"^2 - "b"^2)/"a"`

= `sqrt(1/3 - 1/4)/((1/sqrt(3))`

= `((1/(2sqrt(3))))/((1/sqrt(3))) = 1/2`

∴ ae = `1/sqrt(3) xx 1/2 = 1/(2sqrt(3))`

∴ coordinates of foci = (± ae, 0) = `(± 1/(2sqrt(3)), 0)`.

iii. `"a"/"e" = ((1/sqrt(3)))/((1/2)) = 2/sqrt(3)`

The equations of directrices are

x = `± "a"/"e"`

∴ x = `± 2/sqrt(3)`

iv. Length of latus rectum = `(2"b"^2)/"a"`

= `(2 xx 1/4)/((1/sqrt(3))`

= `sqrt(3)/2`

v. Distance between foci = 2ae

= `2 xx 1/(2sqrt(3)`

= `1/sqrt(3)`

vi. Distance between directrices = `(2"a")/"e"`

= `2 xx 2/sqrt(3)`

= `4/sqrt(3)`

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Conic Sections - Ellipse
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Conic Sections - Exercise 7.2 [पृष्ठ १६३]

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Find the

  1. lengths of the principal axes.
  2. co-ordinates of the focii
  3. equations of directrics
  4. length of the latus rectum
  5. distance between focii
  6. distance between directrices of the ellipse:

3x2 + 4y2 = 12


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  2. co-ordinates of the focii 
  3. equations of directrics 
  4. length of the latus rectum
  5. distance between focii 
  6. distance between directrices of the ellipse:

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