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Maths Conic Sections help

Ogaar

Need help with a question

Reply 1

RDKGames

Study Forum Helper

Original post by Ogaar

Need help with a question

If you substitute the lines equation into the parabola, you will obtain a quadratic equation in x with m as a parameter describing the points of intersection between the line and the curve.

What you aim to prove is equivalent to showing that there is always only one solution to the leftover quadratic you obtain when m is nonzero.

Reply 2

Ogaar

So I square the line equation, substitute it in and I should get a quadratic.
Then, I see if the discriminant is equal to zero which means there is one intersection which means it is a tangent?

Reply 3

RDKGames

Study Forum Helper

Original post by Ogaar

So I square the line equation, substitute it in and I should get a quadratic.
Then, I see if the discriminant is equal to zero which means there is one intersection which means it is a tangent?

Yes

Reply 4

Ogaar

Original post by RDKGames

Yes

I did this and ended up with 14400m^4 -14400m^2... can you identify what I’ve done wrong?

Reply 5

Ogaar

Original post by RDKGames

Yes

Are you still there?

Reply 6

the bear

Original post by Ogaar

Are you still there?

you have put m² instead of m⁴ in the last line

Reply 7

DFranklin

FWIW, if you set X = m^2x then your equation becomes

16X^2-120X+225

(which also equals (4X-15)^2).

Reply 8

Ogaar

Original post by the bear

you have put m² instead of m⁴ in the last line

Thank you, so it’s actually correct as it equals 0!

Reply 9

Ogaar

Original post by DFranklin

FWIW, if you set X = m^2x then your equation becomes

16X^2-120X+225

(which also equals (4X-15)^2).

I got it now, but thanks for the extra help anyway :smile:

Reply 10

DFranklin

Original post by Ogaar

I got it now, but thanks for the extra help anyway :smile:

Having just had a quick play, it's also very viable to do (i) "backwards". That is, show that for m non-zero, there's a tangent to the curve with gradient m and equation y = mx + 15/4m. (I'd say it's actually marginally less work).