Question

The normal to the curve y = x^2+3x+1 at the point (-1,1) cuts the curve again at Q. Find the coordinates of Q.

I think the question is flawed, or I am really retarded...

Answer 1

find the normal first

Answer 2

Its 1 right?

Answer 3

you have a typo in the quesiton i think. the point should be (-1, -1) i think

Answer 4

After m1m2 = -1, where m1=-1

Answer 5

the given point wont even exist on the given curve :o

Answer 6

I knew it! This is a question my teacher gave, I knew it was wrong

Answer 7

good :)
may be solve it for (-1, -1) and tell her the typo...

Answer 8

Thankyou so much sir! Much appreciated!

Answer 9

np :)
yes the slope of tangent at (-1, -1) is 1
so the normal of slope wud be -1

Answer 10

*slope of normal wud be -1

Answer 11

Yep thats true :)

Answer 12

write out the equation of normal and solve it wid the curve

Answer 13

Thankyou so much!

Answer 14

yw :)

Answer 15

Hey Ganesha, once we get the normal how do we solve it with the curve?

Answer 16

you got the normal curve :-
line thru point (-1, -1) wid a slope of -1

y+1 = -1(x+1) ------------------------------(1)

right ?

Answer 17

Yes, and I eventually got 0 = - x - y - 2
What do i do next?

Answer 18

your equation is,

y = x^2+3x+1 -------------------(2)

Answer 19

solve (1) and (2)

that gives u the intersection points of normal and given curve

Answer 20

But for (2) do i use the dy/dx of it or just that equation?

Answer 21

just the equation,
cuz you want to find intersection between curve and normal

Answer 22

Ok, but how do I find the intersection when one equation has power and one doesnt?