# use dy/dx to find the equ. of tangent

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• Oct 10th 2007, 11:53 PM
afeasfaerw23231233
use dy/dx to find the equ. of tangent
bk ex18a p61 q28
find the equations of the two tangents to the curve $y^2 = x^2 y + 2$ at the point where x = 1

http://img3.freeimagehosting.net/uploads/b229bbf16e.jpg
• Oct 11th 2007, 12:13 AM
earboth
Quote:

Originally Posted by afeasfaerw23231233
bk ex18a p61 q28
find the equations of the two tangents to the curve $y^2 = x^2 y + 2$ at the point where x = 1

http://img3.freeimagehosting.net/uploads/b229bbf16e.jpg

Hello,

the "b" of the equation of the tangent is the y-coordinate of the tangent point. The tangent point must be situated on the given curve. Therefore calculate first the coordinates of the tangent point.

You know already that the x-coordinate of the tangent point is x = 1. Plug in this value into the given equation:

$y^2 = 1^2 \cdot y + 2~\implies~y^2-y-2=0~\implies~y=2~\vee~y=-1$

That means you have 2 points which satisfy the conditions: $T_1(1, 2), T_2(1, -1)$

Plug in the values of the coordinates into the equation of the dy/dx to calculate the slope of the tangents.
• Oct 11th 2007, 12:22 AM
earboth
Hello,

in addition to my previous post I've attached a drawing of the tangents and the curve.
• Oct 11th 2007, 12:41 AM
ticbol
Quote:

Originally Posted by afeasfaerw23231233
bk ex18a p61 q28
find the equations of the two tangents to the curve $y^2 = x^2 y + 2$ at the point where x = 1

http://img3.freeimagehosting.net/uploads/b229bbf16e.jpg

Two tangent lines where x = 1?

That means there are two ponts on the curve where x=1.

y^2 = (x^2)y +2
y^2 = (1^2)y +2
y^2 = y +2
y^2 -y -2 = 0
(y -2)(y +1) = 0
y = 2 or -1
So points (1,2) and (1,-1)

You got the dy/dx, which is the slope of the tangent line anywhere on the curve.

dy/dx = 2xy / (2y -x^2)

------------------------------------
At point (1,2)

slope, m1 = (2*1*2) / (2*2 -1^2) = 4/3.

So we have a point and the slope.
Hence, use the point-slope form of the equation of a line,
(y -y1) = m(x -x1)

(y -2) = (4/3)(x -1)
y -2 = (4/3)x -4/3
y = (4/3)x -4/3 +2
y = (4/3)x +(-4 +3*2)/3
y = (4/3)x +2/3 ---------------------one tangent line.

-----------------------------------------
At point (1,-1)

m2 = (2*1*(-1)) / (2(-1) -1^2) = -2/(-3) = 2/3

y -(-1) = (2/3)(x -1)
y +1 = (2/3)x -2/3
y = (2/3)x -2/3 -1
y = (2/3)x -5/3 --------------the other tangent line.
• Oct 11th 2007, 08:42 AM
afeasfaerw23231233
thanks a lot. i misunderstood the question. i thought the the point (1,b) is a point on the tangent which lies outside the circle. but if it is the case, can i find the equation of the tangent?
• Oct 11th 2007, 09:01 AM
earboth
Quote:

Originally Posted by afeasfaerw23231233
thanks a lot. i misunderstood the question. i thought the the point (1,b) is a point on the tangent which lies outside the circle. but if it is the case, can i find the equation of the tangent?

Hello,

which circle? :eek:
The graph of the given equation consists of 2 seprate parts but it dosn't describe a circle.