# Maths Conic Sections help

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#1
Need help with a question
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1 month ago
#2
(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.
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#3
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?
0
1 month ago
#4
(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
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#5
(Original post by RDKGames)
Yes

I did this and ended up with 14400m^4 -14400m^2... can you identify what I’ve done wrong?
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#6
(Original post by RDKGames)
Yes
Are you still there?
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1 month ago
#7
(Original post by Ogaar)
Are you still there?
you have put m2 instead of m4 in the last line
1
1 month ago
#8
FWIW, if you set X = m^2x then your equation becomes (which also equals (4X-15)^2).
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#9
(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!
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#10
(Original post by DFranklin)
FWIW, if you set X = m^2x then your equation becomes (which also equals (4X-15)^2).
I got it now, but thanks for the extra help anyway
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1 month ago
#11
(Original post by Ogaar)
I got it now, but thanks for the extra help anyway
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).
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