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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

Find the equation of the parabola that satisfies the given conditions : Vertex (0,0); focus (–2,0)

$\begin {array} {1 1} (A)\;y^2=8x & \quad (B)\;y^2=-8x \\ (C)\;x^2=8y & \quad (D)\;x^2= -8y \end {array}$

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A)
Toolbox:
  • If the coordinates of the focus is (-a, 0)
  • The equation of the parabola is $y^2=-4ax $ and the curve will be open leftward.
Step 1 :
Given coordinates of the vertex is (0,0) and the coordinates of focus is (-2, 0)
Since the vertex of the parabola is (0,0) and also the focus lies on the negative x - axis.
The axis of the parabola is x - axis.
Hence the equation of the parabola is $y^2=-4ax$
Substituting for $a$ we get,
$y^2=-4(2)x$
$ \Rightarrow y^2=-8x$ is the required equation of the parabola.
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