bk ex18a p61 q28

find the equations of the two tangents to the curve $\displaystyle y^2 = x^2 y + 2 $ at the point where x = 1

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

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- Oct 10th 2007, 10:53 PMafeasfaerw23231233use dy/dx to find the equ. of tangent
bk ex18a p61 q28

find the equations of the two tangents to the curve $\displaystyle y^2 = x^2 y + 2 $ at the point where x = 1

http://img3.freeimagehosting.net/uploads/b229bbf16e.jpg - Oct 10th 2007, 11:13 PMearboth
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:

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

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

Plug in the values of the coordinates into the equation of the dy/dx to calculate the slope of the tangents. - Oct 10th 2007, 11:22 PMearboth
Hello,

in addition to my previous post I've attached a drawing of the tangents and the curve. - Oct 10th 2007, 11:41 PMticbol
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)

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**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, 07:42 AMafeasfaerw23231233
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, 08:01 AMearboth