Question

Equation of a normal at a point on the curve x^2=4y which passes through the point (1,2). Also find the equation of the corresponding tangent.
Please guide me a bit :3

Answer 1

I'd suggest you first find the derivative dy/dx of the given function at (1,2). This represents the SLOPE OF THE TANGENT LINE TO THE CURVE at (1,2). With this info, you can write the equation of the tangent line.

Now note that the NORMAL LINE at (1,2) is perpendicular to the tangent line there. How is the slope of this NORMAL LINE derived from the slope of the tangent line to the curve at (1,2)?

Answer 2

Hint: use the point-slope form of the equation of a straight line:

y-y1 = m(x-x1).

Answer 3

Got it :)
Solving it now :D

Answer 4

Okay, lemme confirm a few things:
slope of the curve is '1'
So slope of the line perpendicular to the slope of the curve should be -1
(since m=-(1/m') for 2 perpendicular lines)
& the equation of the normal should be:
y=3-x
& the equation of the original line, ie: tangent to the curve:
y=x+1
@mathmale is my conclusion right?

Answer 5

shock
its wrong

Answer 6

hold on, i'll get right back to ya

Answer 7

If x^2=4y, then dy/dx = x/2. Therefore, at point (1,2), the slope of the T. L. will be 1/2.

Answer 8

yes yes

Answer 9

i've become ignorant. i mixed up the ordinates

Answer 10

thank you very much for showing your work as you have, and congrats on finding your own error. Nice work! Now, what's the equation of the tangent line?

Answer 11

normal : y=4-2x

Answer 12

original tangent : y=(1/2)(x-3)

Answer 13

Regarding the tangent line: Here's what I'd do with slope=1/2 and point (1,2):

y-2=(1/2)(x-1) => 2y-4=x-1. Does that give us same result as yours?

Answer 14

yes, but its a tangent right? it looks all wrong in the graph maker :O

Answer 15

2y-4=x-1 => 2y=x-1+4 => y = ??

Answer 16

aargh. since when did i mess up addition & subtraction! i am extremely sorry for putting up unnecessary doubts !

Answer 17

its y=(x+3)/2

Answer 18

I really like your attitude. And yes, I agree with that result for the eq'n of the T. L.
Now, what is the slope of the NL at (1,2)?

Answer 19

that should be -2 right?

Answer 20

Yes. Proceed to find the eq'n of the NL.

Answer 21

y=2(2-x)

Answer 22

That's great. At your leisure, try graphing both the TL and the NL on the same set of axes and determine whether or not they intersect at (1,2), as they must. You OK with this?
I need to get off my computer but will wait if you want me to.

Answer 23

hold on

Answer 24

they do intersect, but not looking perpendicular at all. are we still doing something wrong? do get back to it later @mathmale

Answer 25

Don't know whether this will work for you or not, but try this:

http://www.wolframalpha.com/input/?i=y%3D-2x%2B4%2Cy%3D%28x%2B3%29%2F2

If the lines don't look perpendicular, it's almost surely because the scale divisions on the x-and y-axes are different.

Answer 26

I agree: something is wrong here. I'm going to write down this problem and work it out tomorrow morning; I encourage you to do the same. One or both of us will surely find out what went wrong. All 3 curves must intersect at (1,2).

Answer 27

start by rewriting the equation as

\[y = \frac{x^2}{4}\]

then the derivative is

\[y' = \frac{x}{2}\]

find the slope of the tangent at the point (1, 2)

so slope of the tangent is m = 1/2

then find the slope of the normal by finding the negative reciprocal of the tangent slope

\[m_{n} = -2\]

the slope intercept form is y = mx + b

you know m

so you have

y = -2x + b

use the point (1, 2) to find the value of b, and then you'll have the equation of the normal.

hope it makes sense

Answer 28

I think I've discovered what went wrong. My statement that all three curves must go through (1,2) is 'way off. We are to determine a general formula for the normal line to the given curve 4y+x^2. This normal line MUST go through (1,2), which is NOT on the curve (as we/I first thought). So it's back to the drawing board. A very interesting problem!

Answer 29

Note: Point (1,2) is NOT on the curve y = x^2/4

Answer 30

Absolutely right. Thank you for reinforcing that.

Answer 31

Got it! Should I explain?

Answer 32

I'm so pleased that you've obtained the result you were looking for.

Answer 33

Although I'm hitting the sack now, I'll look for your explanation tomorrow morning. Great working with you. Hope to see you again on OpenStudy!

Answer 34

Okay @mathmale I'll just type it down now, do check tom :)

Answer 35

If we start with (a,b) being the point on the curve where the normal drawn will pass through (1,2).
We can find the slope of this normal line as (b-2)/(a-1)
which can be equated to y' at x = a which is a/2

The second equation is (a,b) lies on the curve and therefore, b = a^2/4.

Two equations, two unknowns a and b which can be solved.

Answer 36

Campbell: Agreed, EXCEPT the slope of the nomral will be -2 (not +2).

Answer 37

slope of the normal should be -2/a above.

Answer 38

...where a is the x coordinate of the point at which the normal line intersects the curve, right? and b is the y coordinate, right?

Answer 39

Yes.

Answer 40

and b=(a^2)/4 = y coordinate.

Answer 41

Yes.

Answer 42

Cool. Handshake and salute.

Answer 43

:)

Answer 44

somebody give ranga a medal :D !

Answer 45

@ranga

Answer 46

I mean both of you helped along!!

Answer 47

Thanks sayakdbz.

I am getting a = 2, b = 1

Answer 48

Normal passes through (1,2) and (2,1)
m = (2-1)/(1-2) = -1
Equation is: y - 2 = -1(x - 1)
y - 2 = -x + 1
y = -x + 3

Answer 49

Slope of tangent = 1
Tangent line is: (y - 1) = 1(x - 2)
y = x - 1