# This stumped my teacher... (1 Viewer)

#### allyj104

##### New Member
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...

##### Banned
well A is obviously a negative value...

#### gurmies

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

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

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

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

$\bg_white 100 + 40A = 0$

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

#### Juliaan

##### Member
Winner ~

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

#### gurmies

##### Drover
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

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

#### Juliaan

##### Member
Not 100%,

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