MedVision ad

equation of the tangent to the parabola (1 Viewer)

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
OK, first note that the point (-2, 2) is not on the parabola x<sup>2</sup> = 8y, so there is an error in your question. You might mean: Find the equation of the tangent to the parabola x<sup>2</sup> = 8y at the point where x = -2, in which case the answer is:

Making y the subject of x<sup>2</sup> = 8y, we have y = x<sup>2</sup> / 8
dy/dx = 2x / 8 = x / 4

At x = -2, m<sub>tang</sub> = -2 / 4 = -1 / 2
And, y = (-2)<sup>2</sup> / 8 = 4 / 8 = 1 / 2.

Using the point gradient form of a line, we need the line through (-2, 1 / 2) with gradient -1 / 2. This is:
y - 1 / 2 = (-1 / 2) * (x - -2), which becomes, on multiplication by 2,
2y - 1 = -(x + 2)
ie. x + 2y + 1 = 0, in general form.

Alternately, you may have meant "Find the equation(s) of all tangents to the parabola x<sup>2</sup> = 8y that pass through the point (-2, 2)." This is a harder question.

Let the equation(s) of all such tangent(s) have the form y = mx + b. We know that these tangent(s) pass through (-2, 2), and so:
2 = m(-2) + b
b = 2 + 2m
So, the tangent(s) have the form y = mx + 2m + 2.

These tangent(s) meet the parabola x<sup>2</sup> = 8y, or y = x<sup>2</sup> / 8, when
x<sup>2</sup> / 8 = mx + 2m + 2
x<sup>2</sup> - 8mx - 8(2m + 2) = 0
Since a tangent meets a parabola at only one point, this equation must have only one solution, and so its discriminant is zero.
Thus, (-8m)<sup>2</sup> - 4(1)[-8(2m + 2)] = 0
64m<sup>2</sup> + 64m + 64 = 0
64(m<sup>2</sup> + m + 1) = 0

This equation has no real solution, and so the answer would be that there is no tangent to the parabola x<sup>2</sup> = 8y that passes through (-2, 2).

The fact that there is no answer leads me to suspect that my first interpretation is correct, although other interpretations are possible. For example, you might have meant the point (-4, 2), or the parabola x<sup>2</sup> = 2y, or ...
 
Last edited:

redslert

yes, my actual brain
Joined
Nov 25, 2002
Messages
2,373
Location
Behind You!!
Gender
Male
HSC
2003
Originally posted by wrx
is this yr12 stuff?
i'm not sure

but this kind of question would fall under either locus or diffientiation....

come to think of it NO it is not yr12 but mid year 11 topic!
 

Winston

Active Member
Joined
Aug 30, 2002
Messages
6,128
Gender
Undisclosed
HSC
2003
Yeah yr 11, stuff, it's easy... you just derive the equation, sub in the co-ordinates, and you get your gradient, and then you use the point-gradient formula to find the quation

i.e. y - y0 / x - x0 = gradient
 

Collin

Active Member
Joined
Jun 17, 2003
Messages
5,084
Gender
Undisclosed
HSC
N/A
CM_Tutor, how do you do index notation when posting?
 

shazzam

acta est fabula
Joined
Feb 4, 2004
Messages
260
Gender
Female
HSC
2004
Parametrics?

What, this person should've used parametric equations to work this out, which we certainly do in yr 12 and not yr 11. What the hell, do all you ppl go to James Ruse or s/th?
 

Affinity

Active Member
Joined
Jun 9, 2003
Messages
2,062
Location
Oslo
Gender
Undisclosed
HSC
2003
Originally posted by shazzam
What, this person should've used parametric equations to work this out, which we certainly do in yr 12 and not yr 11. What the hell, do all you ppl go to James Ruse or s/th?
should've? he can do it any way he likes.
And no.. not all of us to go/use to go to james ruse or selectives e.g. me: parramatta high.
 

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
Originally posted by JKDDragon
CM_Tutor, how do you do index notation when posting?
< sup > *insert power* < /sup > with the spaces deleted and < sub > *insert subscript* < /sub > for subscripts. They are just HTML codes so aslong as the forum has it enabled it will work
 

kpq_sniper017

Member
Joined
Dec 18, 2003
Messages
672
&frac12;

i wonder if that worked...??
can you use any html tags in these forums??

lets c:

<font color="#FF0000">Test</font>
 

kpq_sniper017

Member
Joined
Dec 18, 2003
Messages
672
ok cool - it worked.

*for anyone who wants to use fractions instead a/b just type out the word &frac and then the two numbers (numerator first) then a semi-colon.

i think it might only work for certain fractions
 
Last edited by a moderator:

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
Originally posted by pcx_demolition017
&frac12;
But what if you need to do fractions with greater then one digit numerator/denominator eg 13/20 I prefer to improvise, <sup>13</sup>/<sub>20</sub> or <sup>x<sup>2</sup></sup>/<sub>2</sub> try to do that with your frac :p
 
Last edited:

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
Originally posted by Rahul
(x^2)/2
But that defeats the whole purpose of both of them (and you arent pcx), and it gets confusing when you have large equations.

Now when we have a power on the bottom it becomes difficult

<sup>1</sup>/<sub>x<sup>2</sup>+3x-1</sub>

Ok rahul try doing this one neatly

x<sup>x<sup>x<sup>x<sup>x<sup>x<sup>x<sup>x</sup></sup></sup></sup></sup></sup></sup> and x^x^7 doesnt count (although it is right)
 
Last edited:

Rahul

Dead Member
Joined
Dec 14, 2002
Messages
3,647
Location
shadowy shadows
1/(x^2)

it gets annoying, but you can live with it....

how annoying is it to type it out with the sub/sup tags?
 

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
Not that annoying, plus you just copy and paste most of the time because normally it is the same power (ie going from line to line)
 

Collin

Active Member
Joined
Jun 17, 2003
Messages
5,084
Gender
Undisclosed
HSC
N/A
Originally posted by Xayma
< sup > *insert power* < /sup > with the spaces deleted and < sub > *insert subscript* < /sub > for subscripts. They are just HTML codes so aslong as the forum has it enabled it will work
Thank you Xayma.
 

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

Top