maths problem (1 Viewer)

Real Madrid

Member
Joined
Jun 5, 2007
Messages
179
Gender
Male
HSC
2009
The focal chord that cuts the parabola x^2=-6y at (6,-6) cuts the parabola again at X. Find the coordinates at X.

Show all working.
 
Joined
Feb 6, 2007
Messages
628
Location
Terrigal
Gender
Male
HSC
2008
Real Madrid said:
The focal chord that cuts the parabola x^2=-6y at (6,-6) cuts the parabola again at X. Find the coordinates at X.

Show all working.
ok im not gunna do it coz im about to go out
but ill push you in the right direction
find the equation of the focal chord, noting the focus is (0,-3/2) y finding the gradient (m=y2-y1/x2-x1) and using the point gradient formula (y-y1=m(x-x1) )
then when you have it as an expression such as y=12x +3(not the focal chord, just random equation)
sub in the underlined bit into the parabola
then solve for x, should get x=6 and some other value
the sub the other value into either equation for y
 

youngminii

Banned
Joined
Feb 13, 2008
Messages
2,083
Gender
Male
HSC
2009
Real Madrid said:
The focal chord that cuts the parabola x^2=-6y at (6,-6) cuts the parabola again at X. Find the coordinates at X.

Show all working.
So in x^2=-6y, the Focus would be (0,-1/2)
And since the Chord goes through the Focus, we have two points with which we find the Gradient
Therefore m=-11/12

Using Point-Gradient formula, 11x+12y+6=0 is the equation of the Focal Chord
Solving Simultaneously to find the point of intersections between the Parabola and the Focal Chord
We end up with x=-1/2 or x=6
When x=6, y=-6(the Point we were given to start with)
When x=-1/2, y=-1/24
Therefore the point X is (-1/2, -1/24)
Ta-da
 

Danneo

Member
Joined
Sep 12, 2016
Messages
46
Gender
Male
HSC
2018
Lol mate you got the wrong answer though i assume your method was correct, m=-3/4, just sub y=(x^2)/-6 into the tangent equation
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top