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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=10x$

$\begin {array} {1 1} (A)\;focus \bigg(\large\frac{5}{2}, 0\bigg) , Axis : x - axis, Equation \: of \: directrix : 2x+5=0, Length \: of \: the \: latus \: rectum : 10 \\ (B)\; focus \bigg( 0, \large\frac{5}{2} \bigg) , Axis : y - axis, Equation \: of \: directrix : 2y+5=0, Length \: of \: the \: latus \: rectum : 10 \\ (C)\;focus \bigg(-\large\frac{5}{2}, 0\bigg) , Axis : x - axis, Equation \: of \: directrix : 2x-5=0, Length \: of \: the \: latus \: rectum : 10 \\ (D)\;focus \bigg(0, -\large\frac{5}{2} \bigg) , Axis : y - axis, Equation \: of \: directrix : 2y-5=0, Length \: of \: the \: latus \: rectum : 10 \end {array}$

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Toolbox:
  • Equation of a parabola is $y^2=4ax$
  • If x is positive , then the parabola opens towards the right.
  • If the equation involves $y^2$ then the axis of the parabola is x - axis.
  • Equation of the directrix is x = -a or x+a=0
  • Length of the lactus rectum is 4a.
Step 1 :
The given equation is $y^2 = 10x$
Here the coefficient of x is positive.
Hence the parabola opens towards the right.
Comparing this equation with
$y^2=4ax$
$\Rightarrow 4a=10$
$ \Rightarrow a = \large\frac{10}{4} $ or $ \large\frac{5}{2}$
$ \therefore $ Coordinates of the focus = $ \bigg( \large\frac{5}{2}$$, 0 \bigg)$
Since the given equation involves $y^2$ , the axis of the parabola is x - axis.
Axis : x - axis.
Equation of the directrix is
$ x = -\large\frac{5}{2}$ or $x+\large\frac{5}{2}$$=0$
$ \Rightarrow 2x+5=0$
Length of the latus rectum is $ 4 \times \large\frac{5}{2}$$=10$
answered Jul 14, 2014 by thanvigandhi_1
 

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