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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the following : $ y^2=-8x$

$\begin {array} {1 1} (A)\;focus (-2, 0) , Axis : y - axis, Equation \: of \: directrix : x-2=0, Length \: of \: the \: latus \: rectum : 8 \\ (B)\; focus (2, 0) , Axis : x - axis, Equation \: of \: directrix : x+2=0, Length \: of \: the \: latus \: rectum : 8 \\ (C)\;focus (0, -2) , Axis : y - axis, Equation \: of \: directrix : y-2=0, Length \: of \: the \: latus \: rectum : 8 \\ (D)\;focus (0, 2) , Axis : y - axis, Equation \: of \: directrix : y+2=0, Length \: of \: the \: latus \: rectum : 8 \end {array}$

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1 Answer

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Toolbox:
  • Equation of the parabola is $y^2=-4ax$
  • Since the coefficient of x is negative. Hence the parabola opens towards the left.
  • Coordinates of the focus are (-a, 0).
  • Equation of the directrix x= a or x - a = 0
  • Length of the latus rectum is 4a.
Step 1 :
The given equation is $y^2=-8x$
Since the coefficient of x is negative, the parabola opens towards the left.
$ \therefore $ Coordinates of he focus is (-2,0)
Since the given equation involves $y^2$ , the axis is x - axis.
Equation of the directrix is x = 2 or x-2=0.
Length of the latus rectum = $4 \times 2 = 8$
answered Jul 14, 2014 by thanvigandhi_1
 

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