It seems that something is wrong either with this proof or my understanding.

[vkash]

Here i am asking all the things about parabola. I think something is wrong with proof where we prove condition of tangency.
It is done shown(in my book)
y^2=4ax //equation of general parabola
let us assume that line y=mx+c is passing through parabola.
points where it will cut parabola.
(mx+c)^2=4ax
solving this equation.
(mx)^2+x(2mc-4a)+c^2=0
line will touch parabola at one point if of the roots of this equation are equal.
It means it’s discriminant is zero.
(2mc-4a)^2-4m^2c^2=0
=> 16a^2=16amc
a=0; not possible because in that case it will not remain a parabola it became a line y=0.
a=mc
this is the require condition for the line to cut the parabola at one point.
So let's take an example.
y=2. this line cuts the parabola y^2-4x=0 at one point. as we can see on the graphs
but does it obey the equation proved previously.
a=0/2. NO. it is not obeying that equation.
WHY??
the condition is for line to cut the parabola at one point.y=2 is also a line that cuts parabola at one point but not obeying the condition.

 

[Citan Uzuki]

(mx)^2+x(2mc-4a)+c^2=0

To conclude that this equation is a quadratic you need to know that m≠0. If m=0, then this is just a linear equation and will have exactly one solution regardless of what the discriminant is. Since your given line does in fact have m=0, the discriminant need not be zero, and so a is not necessarily equal to mc.

 

[vkash]

To conclude that this equation is a quadratic you need to know that m≠0. If m=0, then this is just a linear equation and will have exactly one solution regardless of what the discriminant is. Since your given line does in fact have m=0, the discriminant need not be zero, and so a is not necessarily equal to mc.

great!
problem solved.
million thanks for answering.

 

[HallsofIvy]

By the way, when m is not 0, "touching the parabola at one point" means "tangent to the parabola" which is what the discriminant being 0 gives.