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1.

If the locus of the point which moves so that the difference (p) 0 of its distance from the points `(5, 0) and (-5,0)` is 2 is `x^2/a^2-y^2/24=1` then a isA. `x^(2)-24y^(2)=1`B. `24x^(2)-y^(2)=24`C. `x^(2)-24y^(2)=24`D. `24x^(2)-y^(2)=1`.

Answer» Correct Answer - B
Discussion
2.

If the parabola `y^(2)=4ax` passes through the point (2,-3) then find the co-ordinates of the focus and the length of latus rectum.

Answer» Correct Answer - Focus `((9)/(8),0)`, latus rectum `=(9)/(2)`
Discussion
3.

Find the vertex, focus, axis , latus rectum and directrix of the parabola `y^(2)+4x+6y+17=0`

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

Find the equation of a parabola whose focus and vertex are (0, 0) and (0, 2) respectively.

Answer» The focus (0,0) and vertex (0,2), of the parabola lie on Y-axis.
Produce SA + AZ
=2+2=4
Draw a perpendicular ZM from Z to the axis of parabola. ZM is the directrix of the parabola whose equation is y-4=0.
Let P(x,y) be any point on the parabola.
Now the equation of parabola
PS=PM
`sqrt((x-0)^(2)+(y-0)^(2))=y-4`
`rArr" "x^(2)+y^(2)=(y-4)^(2)`
`rArr" "x^(2)+y^(2)=y^(2)-8y+16`
`rArr" "x^(2)=-8(y-2)`.
Discussion