# Condition of tangency of a line on a general form parabola

• rajeshmarndi
This is in quadratic form in x i.e it give two value of x i.e we get two intersection point of line on the parabola.But I have a doubt, at this point we also get quadratic in c i.e two y-intercept of the line, why?So to get one intersection point of the line on the parabolawe take (2mc - 2mk -4af

#### rajeshmarndi

<Moderator's note: Moved from a technical forum and thus no template. Effort in post #3.>

What is the condition of tangency of a line y=mx+c on parabola with vertex(h,k) ,say for parabola (y-k)2=4a(x-h)?

I could only find the condition of tangency on standard form of parabola, in the internet?

Thank you.

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I could only find the condition of tangency on standard form of parabola, in the internet?
Why would it matter? It has to have a common point and the derivative a this point has to be the same. This is true for every tangent line. In terms of conditions on m,c,k,a,h: You can derive them based on the definition of tangency.

Is this homework?

Why would it matter? It has to have a common point and the derivative a this point has to be the same. This is true for every tangent line. In terms of conditions on m,c,k,a,h: You can derive them based on the definition of tangency.

Is this homework?
This is how I tried.
A line y = mx+c and a parabola with vertex(h,k) say (y-k)2= 4a (x-h)

I put y = mx+c in (y-k)2 = 4a (x-h)

[ (mx + c) - k ] 2 = 4a (x-h)
(mx + c) 2 + k2 - 2k(mx+c) = 4ax - 4ah
(mx)2 + c2 + 2mxc + k2 - 2kmx -2kc = 4ax - 4ah
(m2) x2 + (2mc - 2km - 4a) x + (c2 -2kc +k2 + 4ah) = 0

This is in quadratic form in x i.e it give two value of x i.e we get two intersection point of line on the parabola. But I have a doubt , at this point we also get quadratic in c i.e two y-intercept of the line, why ?

So to get one intersection point of the line on the parabola
we take (2mc - 2mk -4a) 2 - 4(m2) (c2 -2kc +k2 + 4ah) = 0

4m2 c2 + 4m2 k2 + 16 a2 - 8 m2 ck - 16 mac + 16 mak - 4 m2 c2 + 8 m2 kc - 4m2 k2 - 16 m2 ah = 0

16 a2 - 16 mac + 16 mak - 16 m2 ah = 0
a2 - mca + mka - m2 ah = 0
ah m2 + acm - akm - a2 = 0
(ah) m2 + (ac - ak) m - a2 = 0
h m2 + (c - k) m - a = 0

So again we get two slope(m) of the line y=mx+c , why?

But if I wanted to make one slope, then
(c - k) 2 + 4ha = 0
c2 + k2 - 2ck + 4ha = 0
c2 - 2kc + (k2 + 4ah) = 0

Again I got quadratic in c. So,
4k2 - 4(k2 + 4ah) = 0
4k2 - 4k2 - 16 ah = 0

At last , the condition of tangency of the line y=mx+c is,

ah =0.

I'm sure I am terribly wrong. I also do not know the correct answer.
Please rectify and help me to get the condition of tangency of the line.

This looks a bit messy to me. Sort out what you have:
• a point ##(p,q)## where you want to know the tangent
• a parabola, you choose ##(y-k)^2= 4a(x-h)##
This is a bit unusual, as it looks as if you decided to have a function ##x(y)## which depends on ##y## as free variable. So it's not a function anymore. To make it a function, there are three possibilities
1. Consider ##y## as variable, in which case the straight should be ##x=my+b##, resp. ##y=c##.
2. Chose a branch: positive or negative roots of ##(y-k)^2##, but not both.
3. Change the formula for the parabola and take ##y=ax^2+bx+c##
For the sake of simplicity, I assume you've chosen the third option. So again, what do we have?
• a point ##Q\, : \,(p,q)## where we want to know the tangent
• a parabola ##P\, : \,y=ux^2+vx+w##
• the point is on the parabola, ##Q \in P## or ##q=up^2+vp+w##
• a tangent ##T\, : \,y=mx+b## at ##Q \in T##, i.e. ##q=mp+b## or ##p=c##.
Since we want to know ##T##, and we already have one equation, we simply need another one about ##T.##

Do you know what ##\left. \dfrac{d}{dx}\right|_{x=p} P= \left. \dfrac{d}{dx}\right|_{x=p}(ux^2+vx+w)## geometrically means, resp. can you calculate it?

This is in quadratic form in x i.e it give two value of x i.e we get two intersection point of line on the parabola.
You didn't set any constraints so far, in general a line can intersect a parabola twice. You want it to have only one point in common. You can use this condition (the quadratic equation should have exactly one solution) to get a constraint on the parameters.

I have rewritten my original post and has been shorten.

A line y = mx+c
and a parabola with vertex(h,k) say (y-k)2= 4a (x-h)

I put y = mx+c in
(y-k)2 = 4a (x-h)

it becomes,

[ (mx + c) - k ] 2 = 4a (x-h)
(mx + c) 2 + k2 - 2k(mx+c) = 4ax - 4ah
(mx)2 + c2 + 2mxc + k2 - 2kmx -2kc = 4ax - 4ah
(m2) x2 + (2mc - 2km - 4a) x + (c2 -2kc +k2 + 4ah) = 0

So to get one intersection point of the line on the parabola
we take
(2mc - 2mk -4a) 2 - 4 . (m2) . (c2 -2kc +k2 + 4ah) = 0

4m2 c2 + 4m2 k2 + 16 a2 - 8 m2 ck - 16 mac + 16 mak - 4 m2 c2 + 8 m2 kc - 4m2 k2 - 16 m2 ah = 0

after cancelling +ve and -ve equal term, it reduces to,

16 a2 - 16 mac + 16 mak - 16 m2 ah = 0
a2 - mca + mka - m2 ah = 0
ah m2 + ( c - k ) am - a2 = 0
h m2 + (c - k) m - a = 0
(c - k)m = a - h m2
c - k = a/m - hm
c = a/m - hm + k

Hence c = a/m - hm + k, here c = y-intercept of the line
is the required condition of the line y = mx + c to be a tangent to the above parabola.

This sould be correct, now.

I didn't check every step but the result looks good.
You can test some examples to confirm it as well.

• rajeshmarndi
I didn't check every step but the result looks good.
You can test some examples to confirm it as well.
Thanks..