Math question, help please!?

1, Find the equation of any tangent to the curve representded by x^2-xy+3y^2=132 that is parallel to the straight line defined by x-y=2.

(Answer: x-y-12=0 or x-y+12=0)



2,Find the equations of straight lines throught point A(3,-2) that are tangent to the curve defined by y=x^2-7.

(Answer: 10x-y-32=0 and 2x-y-8=0)





3, Find the equation of the tangent to the curve defined by y=x+ln x that is perpendicular to the line defined by 3x+9y=8.

(Answer: 6x-2y-(2ln 2+2)=0 )



These questions are from the book Harcourt Mathematics 12 (Chpater 9) for high school.



Many thanks!!!Math question, help please!?
1.

x^2-xy+3y^2=132

=%26gt; 2x - y - x dy/dx + 6y dy/dx = 0

For tangent parallel to x - y = 2, dy/dx = 1 and if the tangent having slope 1 is at the point (x', y'),

2x' - y' - x' + 6y' = 0

=%26gt; x' = - 5y'

As (x', y') is on the curve, plugging x' = - 5y' in the eqn. of the curve,

(- 5y')^2 - (- 5y')*y' + 3y'^2 = 132

=%26gt; 33y'^2 = 132 =%26gt; y' = -2 or 2 andcorresponding x' = 10 or - 10

Eqns. of tangents are lines through (-10, 2) and (10, -2) having slope 1 and are

y - 2 = (x + 10) or y + 2 = (x - 10)

i.e., x - y + 12 = 0 or x - y - 12 = 0.



2.

Let the tangents have slope = m

=%26gt; eqn. of tangent is y + 2 = m(x - 3)

Solving with the eqn. of the curve should give two identical roots. Plugging y = m(x - 3) - 2 in the eqn. of the curve

m(x - 3) - 2 = x^2 - 7

=%26gt; x^2 - mx + 3m - 5 = 0

For this eqn. to have two identical roots, its discriminant = 0

=%26gt; m^2 - 4(3m - 5) = 0

=%26gt; m^2 - 12m + 20 = 0

=%26gt; (m - 2)(m - 10) = 0

=%26gt; m = 2 or 10

=%26gt; eqns. of tangents are

y + 2 = 2(x - 3) or y + 2 = 10(x - 3)

i.e., 2x - y - 8 = 0 or 10x - y - 32 = 0



3.

Slope of line 3x + 9y = 8 is - 1/3

Slope of a perpendicular line = 3

For the curve, y = x + ln x

dy/dx = 1 + 1/x = 3

=%26gt; 1/x = 2 and x = 1/2 Corresponding y = 1/2 - ln2

Equation of tangent having slope 3 and passing through

[1/2, 1/2 - ln2] is

y - (1/2 - ln2) = 3 (x - 1/2)

=%26gt; 2y - 1 + 2ln2 = 6x - 3

=%26gt; 6x - 2y - (2ln2 + 2) = 0.