qn (1 Viewer)

[bos1234]

Find the equation of the parabola with vertex the origin, the x-axis is y=0 and the line x-y+3-0 is a tangent to the curve.<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>

 

[kony]

since the parabola has its vertex on the origin, it is in the form x² = 4ay ---- (I)

differentiating,

2x = 4ay'
x = 2ay'
y' = x/2a

now that line, x-y+3=0
y=x+3 ------------- (II)

i.e. y' at the point of contact is 1.

let the point of contact be P (x1, y1)

at P(x1,y1), y' = 1

therefore, x1/2a = 1
therefore, x1 = 2a. ---------- (III)

for intersection P, solving (I) and (II)

x² = 4a(x+3)

at P, x = x1

x1² = 4a(x1+3)

also, x1 = 2a (from III)

so 4a² = 4a(2a + 3)
4a² = 8a² + 12a
a² = 2a² + 3a
a² + 3a = 0
a(a+3) = 0

clearly, a = -3.

therefore, the eqn of the parabola is x² = -12y

woot pretty long proof for a simplish question.

 

[Affinity]

But it could be shorter[think about why this works!]:

we want a such that the equations has 1 double solution for (x,y):
x^2 = 4ay
x-y+3=0

ie the discriminant of:
x - (x^2/4a) + 3 = 0
is 0

(1 + 3/a) = 0
a = -3

 

[kony]

:mad1:
:mad1::mad1:
:mad1::mad1::mad1:
:mad1::mad1::mad1::mad1:
:mad1::mad1::mad1::mad1::mad1:

 

[haque]

how elegant was that-very nice

 

[bos1234]

But it could be shorter[think about why this works!]:

we want a such that the equations has 1 double solution for (x,y):
x^2 = 4ay
x-y+3=0

ie the discriminant of:
x - (x^2/4a) + 3 = 0
is 0

(1 + 3/a) = 0
a = -3

how did you get this step please?
ie the discriminant of:
x - (x^2/4a) + 3 = 0
is 0

 

[kony]

he substituted y=x^2/4a into the equation of the line.

then he used the fact that, if the there is only one intersection between a parabola and a line, i.e. a tangeant, then the discriminant = 0. (this applies for not just parabolas, but all 2nd-degree curves, i.e. conics and probably various others that are beyond the hsc course)

 

[bos1234]

y is discriminant 0 when there is a tangent??

 

[Affinity]

if the discriminant is positve you get 2 solutions, ie the line cuts the parabola (which means it's not a tangent because all tangents to a parabola meets it only once)
if the discriminant is negative it means there are no solutions, again not a tangent.
so when the line is a tangent, the discriminant is 0

 

[Riviet]

Just adding to what Affinity said, when the discriminant is zero, you get one solution:

x = { -b + sqrt(discriminant) } / 2a

= (-b + 0)/2a

= -b/2a

Since a and b are constants, there is only one solution and therefore one point of intersection => tangent.

 

[bos1234]

ahh kk got it nw.. so discriminant doesnt always mean roots with x-axis..

 

[jyu]

Alternative parabola: y^2 = 12x

:wave:

 

[Affinity]

smartass