# normal line?

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• Oct 11th 2006, 09:41 PM
cyberdx16
normal line?
alright heres the question:

the point (1,2) lies on the curve described by the equ. x^2-xy+y^2=3
find the point other than (1,2) where the normal line to the curve at (1,2) crosses the curve.

ok i really dont know what there asking here... i mean i know how to find the normal line of the tangent line, neg recipricale(sp) of dy/dx x^2-xy+y^2=3
but what do they mean by find the point other than (1,2) where the normal line to the curve at (1,2) crosses the curve.
how do i find that?

do they mean find a equ of a line... meaning find the normal line from the given pt and equ?
• Oct 11th 2006, 11:36 PM
earboth
Quote:

Originally Posted by cyberdx16
alright heres the question:

the point (1,2) lies on the curve described by the equ. x^2-xy+y^2=3
find the point other than (1,2) where the normal line to the curve at (1,2) crosses the curve.

ok i really dont know what there asking here... i mean i know how to find the normal line of the tangent line, neg recipricale(sp) of dy/dx x^2-xy+y^2=3
but what do they mean by find the point other than (1,2) where the normal line to the curve at (1,2) crosses the curve.
how do i find that?

do they mean find a equ of a line... meaning find the normal line from the given pt and equ?

Hi,

your equation describes an ellipse with its centre in the origin.

At (1,2) the tangent to the ellipse has the slope zero, thus the normal line is a parallel to the y-axis. The other point which you were looking for has the same x-value as the given point. Plug in x = 1 into your equation and solve for y. You'll get y = 2 (that's the given point) or y = -1.

So the unknown point is (1, -1).

EB
• Oct 11th 2006, 11:44 PM
earboth
Quote:

Originally Posted by cyberdx16
alright heres the question:

the point (1,2) lies on the curve described by the equ. x^2-xy+y^2=3
find the point other than (1,2) where the normal line to the curve at (1,2) crosses the curve....

Hi,

it's me again. I've attached a diagram to illustrate my previous post.
• Oct 12th 2006, 12:21 PM
cyberdx16
still a little lost?