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Q)

Find the axies, vertex, focus, equation of directrix , latus rectum , length of the latus rectum for the following parabolas and hence sketch their graphs. $y^{2}+8x-6y+1=0$.

This is the fourth part of the multi-part question Q2

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A)
Toolbox:
  • Standard parabolas :
  • (i)$y^2=4ax$
  • In this case the axis of symmetry is $x$-axis.
  • The focus is $F(a,0)$ and the equation of the directrix is $x=-a$
  • (ii)$y^2=-4ax$
  • In this case the focus is $F(-a,0)$ and the equation of directrix is $x=a$
  • (iii)$x^2=4ay$
  • In this case the axis of symmetry is $y$-axis.
  • The focus is $F(0,a)$ and the equation of directrix is $y=-a$
  • (iv)$x^2=-4ay$
  • In this case the focus is $F(0,-a)$ and the equation of directrix is $y=a$
  • Shifting of origin from (0,0) to $(h,k)$ then if the new axes form $X-Y$ coordinate system then
  • $x=X+a$
  • $y=Y+k$
  • The new coordinates of a point $P(x,y)$ become $X=x-a$,$Y=y-k$.
  • General forms of the parabola when the origin is shifted to $(h,k)$ and the new axes are parallel to the $xoy$ axes.
  • Equations with respect to $xoy$ axes :
  • $(y-k)^2=4a(x-h)$
  • $(y-k)^2=-4a(x-h)$
  • $(x-h)^2=4a(y-k)$
  • $(x-h)^2=-4a(y-k)$
Step 1:
$y^2+8x-6y+1=0$
$y^2-6y+1=-8x$
$y^2-6y+1+8=-8x+8$
$(y-3)^2=-8(x-1)$
Shifting the origin to $(1,3)$ by translation of axes
$X=x-1$
$Y=y-3$
Step 2:
$XY$ axis :
$y^2=-8x$
One parabola opens out towards the left.
$4a=8$
$a=2$
Axis : $Y=0$
Vertex : $(0,0)$
Focus : $(-2,0)$
Equation of directrix : $X=2$
Equation of LR : $X=-2$
$LR=4a=8$
$\Rightarrow 4\times 2=8$
Step 3:
$xy$ axis :
Axis : $y-3=0$
Vertex : $(1,3)$
Focus : $(-1,3)$
Equation of directrix : $x-1=2\Rightarrow 3$
Equation of LR : $x-1=-2\Rightarrow -1$
$LR=8$
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