conic section

[Leons]

1. One point P at parabola $$(x-a)^2=4ay$$ have parameter coordinate
$$x= a+2at$$ , $$y=at^2$$a) find an equation of tangent and normal to the parabola at point P

 

[Harry_the_cat]

Where are you stuck? Are you given a specific point P?

 

[Leons]

Where are you stuck? Are you given a specific point P?

I stuck after I find the slope. The question does not given a specific point P.
this is the slope that I get : $$(x-a)/2a$$

 

[Dr.Peterson]

I stuck after I find the slope. The question does not given a specific point P.
this is the slope that I get : $$(x-a)/2a$$

As I read the problem, the point P is specified by a parameter t, namely P $(a+2at, at^2)$.

You are to find the slope at that point; your slope is correct as a function of x, so replace x with $a+2at$. (The result is interesting!) Then find the equation of the line with that slope through that point.

 

[Leons]

As I read the problem, the point P is specified by a parameter t, namely P $(a+2at, at^2)$.

You are to find the slope at that point; your slope is correct as a function of x, so replace x with $a+2at$. (The result is interesting!) Then find the equation of the line with that slope through that point.

can you check whether my answer is correct or not?

slope when x=a+2at
$$\frac{a+2at-a}{2a}=t$$slope=t

tangent equation,$$(y-y_1)=m(x-x_1)$$$$(y-at^2)=t(x-(a+2at))$$$$y=tx-at-2at^2+at^2$$$$y=tx-at-at^2$$

 

[Dr.Peterson]

can you check whether my answer is correct or not?

slope when x=a+2at
$$\frac{a+2at-a}{2a}=t$$slope=t

tangent equation,$$(y-y_1)=m(x-x_1)$$$$(y-at^2)=t(x-(a+2at))$$$$y=tx-at-2at^2+at^2$$$$y=tx-at-at^2$$

Yes, that's correct. As confirmation, I've graphed the parabola and the line:

Here, t is the parameter for drawing the curve, and s is a particular value of t to determine the point.

 

[Leons]

Yes, that's correct. As confirmation, I've graphed the parabola and the line:
View attachment 32761
Here, t is the parameter for drawing the curve, and s is a particular value of t to determine the point.

thank you so much

 

[lookagain]

That was one hairy conic, Junior.

Excuse me, I mean Leons.

 

[topsquark]

That was one hairy conic, Junior.

Excuse me, I mean Leons.

You're dating yourself with that one!

-Dan

 

[Leons]

That was one hairy conic, Junior.

Excuse me, I mean Leons.

hairy conic??

 

[topsquark]

hairy conic??

See Harry Connick, Jr.

-Dan

 

[Leons]

See Harry Connick, Jr.

-Dan

i still dont understand?.. what does relate with him

 

[Leons]

1. One point P at parabola $$(x-a)^2=4ay$$ have parameter coordinate
$$x= a+2at$$ , $$y=at^2$$a) find an equation of tangent and normal to the parabola at point P

i already got the equation of tangent and parabola which are
$$y=tx-at-at^2$$$$ty+x=a+2at+at^3$$
now i stuck at this question ,,


if the normal and tangent line across x-axis at point T and N respectively, show that
$$\frac{PT^2}{TN}=at$$

 

[Dr.Peterson]

now i stuck at this question ,,

if the normal and tangent line across x-axis at point T and N respectively, show that
$$\frac{PT^2}{TN}=at$$

Please show whatever you have tried for this part. The points T and N are the x-intercepts of your two lines. How do you find those?

 

[Leons]

Please show whatever you have tried for this part. The points T and N are the x-intercepts of your two lines. How do you find those?

tangent line at point N
$$0=tx-at-at^2$$Normal line at point T
$$0=a+2at+at^3-x$$

 

[Leons]

tangent line at point N
$$0=tx-at-at^2$$Normal line at point T
$$0=a+2at+at^3-x$$

im a bit confused, whether i have to prove on the left hand side or right hand side

 

[Dr.Peterson]

tangent line at point N
$$0=tx-at-at^2$$Normal line at point T
$$0=a+2at+at^3-x$$

im a bit confused, whether i have to prove on the left hand side or right hand side

Prove what?

The goal is to solve each equation for x, right? Do that.

 

[Leons]

Prove what?

The goal is to solve each equation for x, right? Do that.

okay i will try solve that, please check for me
Point T:
$$x=\frac{at+at^2}{t}$$Coordinate point T :$$(\frac{at+at^2}{t},0)$$Point N:
$$x=a+2at+at^3$$Coordinate point N
$$(a+2at+at^3,0)$$Coordinate point P
$$(a+2at,at^2)$$
distance PT:
$$\sqrt{(a+2at-(\frac{at+at^2}{t}))^2+(at^2-0)^2}$$$$\sqrt{a^2t^2+a^2t^4}$$$$PT^2=a^2t^2+a^2t^4$$
distance TN:
$$\sqrt{(\frac{at+at^2}{t}-(a+2at+at^3))^2+(0-0)^2}$$$$TN=-at-at^3$$
so,
$$\frac{PT^2}{TN}=\frac{a^2t^2+a^2t^4}{-at-at^3}=\frac{a^2t^2(1+t^2)}{-at(1+t^2)}=-at$$
i got -at,, what i know distance should be positive, so i want to ask whether i can change the sign or not?

 

[Dr.Peterson]

i got -at, what i know distance should be positive, so i want to ask whether i can change the sign or not?

Since a distance is always positive, your expression for TN should have had an absolute value. (The square root of a square is the absolute value.)

So, yes, your final answer will be at, if a and t are positive! Was that stated in the problem? If not, the answer should really be $|at|$.

Also, since N is to the right of T when $t>0$, you might have subtracted T from N rather than N from T in your calculation, to avoid the absolute value. Here is a graph, for $t=1.6$:

​

 

[Leons]

Since a distance is always positive, your expression for TN should have had an absolute value. (The square root of a square is the absolute value.)

So, yes, your final answer will be at, if a and t are positive! Was that stated in the problem? If not, the answer should really be $|at|$.

Also, since N is to the right of T when $t>0$, you might have subtracted T from N rather than N from T in your calculation, to avoid the absolute value. Here is a graph, for $t=1.6$:

View attachment 32791​

oh,, now i understand. Thank you so much !!