Condition of tangency of a line on a general form parabola

[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.

 

[mfb]

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?

 

[rajeshmarndi]

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.

​

 

[fresh_42]

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?

 

[mfb]

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.

 

[rajeshmarndi]

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.

 

[mfb]

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..