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Grade 12th passMost Scoring Topics in IIT JEE

Qno 1
The points on the parobola y^2 = 8x whose focal distance is 4
q no 2
The parabola passes through (0,4) (1,9) and (-2,6) and its axis is parallel toy axis then its equation
Qno 3
The equation of the parabola with latus rectum joining the points (6,7) and (6,-7) is
q no 4
If the focus is (1,-1) and the directrix is the line x+2y-9=0 the vertex of the parabola is

Profile image of anila
9 Years agoGrade 12th pass
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1 Answer

Profile image of Sayam
9 Years ago
Q1--y2=8x so a=2
focal distance =a(1+t2)=4......i
solving t, you will get t=1 or -1. The parametric point is of the form (at2,2at).
answer is (2,4) and (2,-4)
Q2-- AS the parabola is parallel to y axis, so it will be of the form ax2 +bx+c=y
you are given three points. Substitute them in the above equation. Solve them. you will get a=2,b=3,c=4
the required equation is 2x2 +3x+4=y. you can also make this it in standard form.
Q3--
the points of lactus rectum is (6,7) and (6,-7). focus will be its mid point. focus is (6,0).
length of lactus rectum is=4a=14 implies a=3.5. So vertex of parabola is (2.5,0)
the required equation is y2=14(x-2.5)
Q4-- focus is (1,-1) and the directrix  x+2y-9=0. find the foot of perpendicular of focus on directrix.
it comes out to be (3,3). vertex will be mid point of (1,-1) and (3,3). answer is (2,1) .. hope this clears your doubt.

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