1. Given that the straight line y=c-3x does not intersect the curve xy=3, find the range of values for c.
2. Find the range of values for c for which the line y=cx+6 does not meet the curve 2x^2-xy=3.
3. Find the range of values for k for which 8y=x+2k is a tangent to the curve 2y^2=x+k
Bruh i don't understand yallz explanation
The slope of line cannot be that of curve or greater.
slope line=-3
slope curve>slope line
-3<-3/x^2
x^2>1
x>1
check x=10 slope line=-3
slope curve=-3/100=-.03
check x=1/2 slope curve=-12
do the others the same way
number one is supposed to be -6<c<6
number two is supposed to be c>5
and number three is supposed to be k=8
the slope of xy=3 is y' = -3/x^2
So, the slope is -3 at x=1
The tangent to the curve at x=1 is
y-3 = -3(x-1)
y = -3x+6
y = 6-3x
so, for y = c-3x, if c < 6, the line falls below the curve. Similarly, for x<0, the line lies above the other branch of the curve if c > -6.
So, -6 < c < 6
Follow this logic for the other parts
1. Well, let's find where the straight line and the curve intersect. We can substitute y=c-3x into the equation xy=3 and get (c-3x)x=3. This gives us a quadratic equation c*x - 3x^2 - 3 = 0. For this equation to have no real solutions, the discriminant must be negative. So, we have (0)^2 - 4(c)(-3) < 0. Simplifying this gives us 12c < 0, which means c < 0. Therefore, the range of values for c is c < 0. Why negative? Well, lines and curves can be moody sometimes. They just don't like each other in certain neighborhoods.
2. Let's do some detective work here. Substitute y=cx+6 into the equation 2x^2-xy=3, and we have 2x^2 - x(cx+6) = 3. Simplifying this equation, we get 2x^2 - c*x^2 - 6x - 3 = 0. For this equation to have no real solutions, the discriminant must be negative. So, we have (0)^2 - 4(2-c)(-3) < 0. Solving this inequality gives us -20 + 12c < 0, which means c < 20/12, or c < 5/3. Therefore, the range of values for c is c < 5/3. Just like two puzzle pieces that can't fit together, some lines and curves just won't meet.
3. Ah, the mysterious world of tangents. To find the range of values for k, we need to find where the line 8y = x + 2k is tangent to the curve 2y^2 = x + k. Let's find the derivative of the curve and set it equal to the slope of the line, which is 8. So, differentiating 2y^2 = x + k gives us 4y(dy/dx) = 1. Rearranging this equation gives us dy/dx = 1/(4y). Setting this equal to the slope of the line, we get 1/(4y) = 8. Simplifying this equation gives us y = 1/32. Substituting this back into the equation 2y^2 = x + k, we get 2(1/32)^2 = 1/32 + k. Solving this equation gives us k = 1/2048. Therefore, the range of values for k is k = 1/2048. Remember, tangents are elusive creatures.
To find the range of values for the constant c in the given equations, we need to consider the points of intersection between the line and the curve. When there are no intersection points, it means the line does not meet the curve.
1. To find the range of values for c in the equation y = c - 3x not intersecting the curve xy = 3:
a) Substitute y = c - 3x into the equation xy = 3:
(c - 3x)x = 3
b) Rearrange the equation:
x^2 - (c/3)x + 3 = 0
c) For no intersection, the discriminant of the quadratic equation should be less than zero:
(c/3)^2 - 4(1)(3) < 0
c^2 - 36 < 0
d) Solve the inequality:
c^2 < 36
-6 < c < 6
Therefore, the range of values for c is -6 < c < 6.
2. To find the range of values for c in the equation y = cx + 6 not meeting the curve 2x^2 - xy = 3:
a) Substitute y = cx + 6 into the equation:
2x^2 - x(cx + 6) = 3
b) Rearrange the equation:
2x^2 - cx^2 - 6x - 3 = 0
c) For no intersection, the discriminant should be less than zero:
(-c)^2 - 4(2)(-6 - 3) < 0
c^2 + 72 < 0 (Note: The negative sign is squared since we want the discriminant to be negative)
d) Solve the inequality:
c^2 < -72
Since the square of a real number cannot be negative, there are no values for c that satisfy the inequality. Hence, the line will always meet the curve.
3. To find the range of values for k in the equation 8y = x + 2k being a tangent to the curve 2y^2 = x + k:
a) Differentiate the curve equation with respect to y:
4yy' = 1 + 0 (since the derivative of k with respect to y is 0)
b) Simplify the equation to get y' in terms of y:
y' = 1/(4y)
c) Substitute the equation of the line into the derivative equation:
8(1/(4y)) = 1
d) Simplify the equation:
2/y = 1
y = 2
e) Substitute the value of y back into the line equation:
8(2) = x + 2k
16 = x + 2k
x = 16 - 2k
Since the line is a tangent to the curve, it means there is one common point between the line and the curve. Substitute this value into the curve equation:
2(2)^2 = 16 - 2k + k
8 = 16 - k
k = 16 - 8
k = 8
Therefore, the range of values for k is k = 8.