Find the vertex, focus, axis , latus rectum and directrix of the parab

1.	Find the vertex, focus, axis , latus rectum and directrix of the parabola `y^(2)+4x+6y+17=0`
Answer» Equation of parabola<br> `y^(2)+4x+6y+17=0`<br> `rArr" "y^(2)+6y+9=-4x-17+9`<br> `rArr" "(y+3)^(2)=-4(x+2)`<br> `rArr" "Y^(2)=-4X`<br> Comparing with `Y^(2)=-4aX`<br> 4a=4<br> `rArr" "a=1`<br> Vertex A = (0,0)<br> `rArr" "X=0,Y=0`<br> `rArr" "x+2=0,y+3=0`<br> `rArr" "x=-2,y=-3`<br> `:.` Co-ordinates of vertex = (-2,-3).<br> Focus X = -a,Y=0<br> `rArr" "x+2=-1,y+3=0`<br> `rArr" "x=-3,y=-3`<br> `:.` Co-ordinates of focus = (-3, -3).<br> Equation of axis Y=0<br> `rArr" "y+3=0`.<br> Equation of directrix X=a<br> `rArr" "x+2=1`<br> `rArr" "x+1=0`<br> Length of latus rectum = 4a = 4.

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