IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

hary

the equation of the parabola having focus (-2,3) n vertex (1,2)

pls mention any shortcut meth time alloted to this q is 67 sec

hary

plssssssss reply!!!!!!!

hary

pls help its urgent!!

hary

common plsssssssss solve it!!!!!

arpan1

EDITED

hary

hey da axis of parabola is not parabola is not parallel to da x axis u didnt notice dat da standard eq y^2=4ax is not valid here

ashish_banga

is the answer (y-2)^2 = - 12(x-1)

hary

nope da ans is x^2+6xy+9y^2+106x-82y+9=0
here da ans is not parallel to x axis notice dat point

rachit12

the technique here is the distance btw focus nd on a pt on parabola is equal to the the pt on parabola nd vertex !!

jst equate the two distances u'll gt the answer !!!

hary

how is dat ?? fr a parabola dist bet a pt on da parabola n da focus n dist bet pt on da parabola n directrix r equal right ????

arpan1

START WITH THE DEFINITION THEN

EQN OF PARABOLA

(X + 2)²+ (Y-3)² = (3Y - X + C )² /10

THIS IS THE REQUIRED EXPRESSION

RATE ME

hary

but how to find c???

joyfrancis

the vertex is always the mid pt of the line joining the focus and the pt of intersection of the directric and the axis of the parabola.

so by midpoint formula , the point whr the directrix cuts the axis is (4,1)

so by the definition of a parabola..the distance of any pt on it from the focus is equal to the distance from directrrix..

slope of axis = -1/3..so slope of directrix=3

by slope pt form eqn of directrix = 3x-y-11=0

so eqn of parabola is

(x+2)^2 + (y-3)^2 = ((3x+y-11)^2)/10..now expand to get the exact answer..

arpan1

WELL c CAN B FOUND BY VARIOUS METHODS ONE OF WHICH IS MENTIONED ABOVE

hary

thanx guys got it!!

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