This stumped my teacher... (1 Viewer)

allyj104 · Feb 23, 2009

Ok, i kinda need help with this stupid quadratic/parabola/locus question

A parabola has its equation in the form y=Ax^2 where A is constant. The line y=10x+10 is a tangent to the parabola y=Ax^2. Find the value of A.

Good Luck...

shady145 · Feb 23, 2009

well A is obviously a negative value...

gurmies · Feb 23, 2009

$y = Ax^{2} \,\,\,[1]$

$y = 10x + 10 \,\,\, [2]$

$Ax^{2} - 10x - 10 = 0 \,\,\, [1] - [2]$

$If \,\, tangent, \,\, \Delta = 0$

$100 + 40A = 0$

$A = -\frac{5}{2}$

Juliaan · Feb 23, 2009

Winner ~

Yet i dont know what you did after you said, If(?) Tangent, ^ = 0

gurmies · Feb 23, 2009

Understand that if two curves are equal to eachother, you can solve simultaneously like I did. A tangent you'll find, is a special case where when you form the equation like I had, there is only one solution, as that's the definition of a tangent - only one point of contact. Using this fact, I link it to the fact that in any quadratic equation, when there's only one solution, the discriminant is equal to 0. This allows you to find a numerical value for A.

allyj104 · Feb 23, 2009

Thanku kindly good sir, u have outdone 2 of my schools maths teachers... thats very hard 2 do *doffs hat*

Juliaan · Feb 23, 2009

Not 100%,

Though, .. (i tried to explain what i thought you were saying, though i realised i'd look like an idiot luls~)

This stumped my teacher... (1 Viewer)

allyj104

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shady145

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gurmies

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Juliaan

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