Exploring Tangents on Parabolas: A Comparison of y^2=4ax and x^2=4ay

[fireflies]

I know a tangent drawn on parabola having equation like
y^2=4ax is
y=mx+(a/m)
which provides c=a/m
Then how is it going to turn for equation like x^2=4ay?

From my derivation it will be like -c=am^2
when the equation of tangent is y=mx+c.

The derivation comes from the following:
y=mx+c
or, x=(y/m)-(c/m)

So, comparing with the tangent on
y^2=4ax we get

-(c/m)=a/(1/m)
that is -c= am^2

But the problem arises when in a question saying find the common tangent on y^2=4ax and x^2=4ay, the solution was made taking the tangents for each parabola as
y=mx+a/m and
x= my + a/m respectively.

Shouldn't the later one be x=(y/m)-(c/m)
i.e x=(y/m)-am ?

 

[HallsofIvy]

What you say,
"I know a tangent drawn on parabola having equation like
y^2=4ax is
y=mx+(a/m)
which provides c=a/m"

Simply doesn't mean anything because you have not said what "m" means!

Please rewrite this, telling us what "m" is, so that it makes sense.

 

[Mentallic]

Simply doesn't mean anything because you have not said what "m" means!

Please rewrite this, telling us what "m" is, so that it makes sense.

If choosing various m gave all the tangents to the parabola, then it'd make sense. This is not the case however.

 

[fireflies]

It's a general straight line equation, where m is the slope of a tangent

 

[fireflies]

If choosing various m gave all the tangents to the parabola, then it'd make sense. This is not the case however.

If choosing various m gave all the tangents to the parabola, then it'd make sense. This is not the case however.

Yes, it is the case.

If the tangent is y=mx+c

where m=slope of the line
c=intercept it cuts on y axis

And it is a tangent to a general parabola of equation

y^2=4ax

then c=a/m

 

[Mentallic]

Yes, sorry, you're right. I made a quick base-case graph check and it went wrong somewhere along the way.

The derivation for your new tangent problem is very simple to derive. Since you've already correctly found that
$y=mx+a/m$ is tangent to $y^2=4ax$
then symmetrically,
$x=my+a/m$ is tangent to $x^2=4ay$

So, comparing with the tangent on
y^2=4ax we get

-(c/m)=a/(1/m)
that is -c= am^2

You'll need to justify this last step. I can't follow that line of thought.

 

[fireflies]

You'll need to justify this last step. I can't follow that line of thought.

In this line you are taking x=my+c as a tangent of the x^2=4ay

Here, m is not the slope, c is not the intercept of y-axis.

Here m= 1/(slope) and
c=-(intercept/slope)

I just put in case of m, 1/m and in place of c, -(c/m) according to the conventional meaning of m and c.

I just tried it on paper. Both are actually same, denoting the same meaning until you are confused what m and c means in which equation.