InterviewSolution
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InterviewSolution
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Find the vertex, focus, equation of axis, equation of axis, equation of directrix, length of latus rectum and the co-ordinates of the ends of latus rectum for the following parabolas : (i) `y^(2) =8x` (ii) `y^(2) = -12x` (iii) `x^(2)=6y` (iv) `x^(2)=-2y`. |
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Answer» (i) `y^(2) = 8x` Comparing with `y^(2) =4ax` `4a = 8 rArr a=2` `:.` Co-ordinates of vertex = (0, 0) Co-ordinates of focus `= (alpha,0)=(2,0)` Equation of axis `y=0` Equation of directrix `x = -a` `rArr x=-2` Length of latus rectum `= 4 alpha =8` Co-ordinates of the ends of the latus rectum `= (a, pm 2a)` `=(2, pm 4)` (ii) `y^(2) = -12x` Comparing with `y^(2)=-4ax` `4a = 12 rArr a=3` `:.` Co-ordiantes of vertex = (0,0) Co-ordiantes of focus `=(-a,0)=(-3,0)` Equation of axis `y=0` Equation of directrix `x = a rArr x =3` Length of latus rectum `= 4a =12` Co-ordinates of the ends of the latus rectum `= (-a, pm 2a)` `=(-3, pm 6)`. (iii) `x^(2)=6y` Comparing `x^(2)=4ay` `4a=6" "rArr" "a=(3)/(2)` `:.` Co-ordinates of vertex = (0,0). Co-ordainates of focus = `(0,a)=(0,(3)/(2))`. Equation of axis x=0. Equation of latus rectum y = - a `rArr" "y=-(3)/(2)`. Length of latus rectum = 4a=6. Co-ordinates of the ends of the latus rectum `(pm2a,a)=(pm3,(3)/(2))`. (iv) `x^(2)=-2y` Comparing with `x^(2)=-4ay` `4a=2" "rArr" "a=(1)/(2)` `:.` Co-ordinates of vertex = (0,0) Co-ordinates of focus `=(0,-a)=(0,-(1)/(2))` Equation of axis x = 0. Equation of directrix = a. `rArr" "y=(1)/(2)` Length of latus rectum = 4a = 2. Co-ordinates of the ends of latus ratus rectum `=(pm2a,-a)=(pm1,(1)/(2))`. |
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