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parabola equation

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parabola equation

by happymanocha » Mon Mar 19, 2012 6:25 pm
The equation x = 2 square(y) + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is


A : 3x + y + 15 = 0
B : y - 3x - 11 = 0
C : -3x + y - 16.5 = 0
D : -2x - y - 7 = 0
E : -3x + y + 13.5 = 0
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by killer1387 » Mon Mar 19, 2012 6:33 pm
happymanocha wrote:The equation x = 2 square(y) + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is


A : 3x + y + 15 = 0
B : y - 3x - 11 = 0
C : -3x + y - 16.5 = 0
D : -2x - y - 7 = 0
E : -3x + y + 13.5 = 0
C

line L; y=3x+c
now L intersects parabola at x=-5 and y=3/2(from parabola equation)

hence c=3/2-3*(-5)=16.5

hence L: y=3x+16.5

C
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by spartacus1412 » Tue Mar 20, 2012 12:03 am
the parabola 2sq(y)+5y-17 = x has two solutions for x = -5.
(-5, -4) and (-5, 3/2) both satisfy the equation of the parabola.

the equation of the line is y = mx +c
for x= -5 amd m =3(given)the equation is

y = 3(x)+c

since the line and parabola meet in the upper left quadrant (the second quadrant hwere both x and y are negative) the possible point of intersection is (-5, -4).

Hence, line is :
-4 = 3(-5)+c
-4 = -15 +c
c = 11
line is y = 3x +11
y-3x-11 =0.
option B.
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by coolhabhi » Tue Mar 20, 2012 12:02 pm
spartacus1412 wrote:the parabola 2sq(y)+5y-17 = x has two solutions for x = -5.
(-5, -4) and (-5, 3/2) both satisfy the equation of the parabola.

the equation of the line is y = mx +c
for x= -5 amd m =3(given)the equation is

y = 3(x)+c

since the line and parabola meet in the upper left quadrant (the second quadrant hwere both x and y are negative) the possible point of intersection is (-5, -4).

Hence, line is :
-4 = 3(-5)+c
-4 = -15 +c
c = 11
line is y = 3x +11
y-3x-11 =0.
option B.
All your calculation is correct but for a mistake...
since the line and parabola meet in the upper left quadrant (the second quadrant hwere both x and y are negative) the possible point of intersection is (-5, -4).

In the second Quadrant X is -ve and Y is +ve.
So we should take (-5, 3/2) as our point. Then

Hence, line is :
3/2 = 3(-5)+c
3/2 = -15 +c
c = 33/2
c = 16.5
line is y = 3x +16.5
y-3x-16.5 =0.
option C
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by ronnie1985 » Wed Mar 21, 2012 9:36 am
By solving the parabola eqn, y = -4 or 3/2. Upper left quadrant => x<0 and y>0. Therefore y = 3/2.

y = mx+c is eq of the line. Solving for c we get c = 33/2 = 16.5. Therefore equation of line is y = 3x+16.5. (C)
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