f(x)=kx^2+3

a) if the tangent lines to the graph of f at (t,f(t)) and (-t, f(-t)) are perpendicular, find t in terms of k
b) Find the coordinate of the point of intersection of the tangent lines mentioned in part (a) in terms of k

Start with the given parabola:

f(x)=kx²+3
The derivative is
f'(x)=2kx
(a)
We are given that the tangent lines at
(t, f(t)) and (-t,f(-t)) are perpendicular, which means that the product of the slopes =-1, or
f(t)*f(-t)=-1
2kt*2k(-t)=-1
Solving for t, we get
t=1/(2k) ....(1)

(b)
The slope of each tangent is given by
m1=f'(t)=2k/(2k)=1, ...(2) and
m2=f'(-t)=2k/(-2k)=-1 ...(3)

f(t)=k(t²)+3=1/(4k) + 3 ...(4)
f(-t)=k(-t)²+3=1/(4k) + 3 ...(5)

The equations of the tangent lines are therefore:
(y-f(t))=m1(x-t) ...(6)
(y-f(-t))=m2(x-(-t)) ...(7)

Solve (6) and (7) for x and y.
Hint: x=0 by symmetry.

a) Well, if the tangent lines to the graph of f at (t, f(t)) and (-t, f(-t)) are perpendicular, it means their slopes are negative inverses of each other. The slope of the tangent line can be found by taking the derivative of f(x). So, let's find the slopes of the tangent lines:

For the point (t, f(t)):
The derivative of f(x) = kx^2 + 3 is f'(x) = 2kx. So, the slope of the tangent line at (t, f(t)) is f'(t) = 2kt.

For the point (-t, f(-t)):
The derivative of f(x) = kx^2 + 3 is f'(x) = 2kx. So, the slope of the tangent line at (-t, f(-t)) is f'(-t) = 2k(-t) = -2kt.

Since the slopes are negative inverses of each other, we have 2kt = -1/(-2kt). Solving this equation, we get:

2kt = 1/2kt
4k^2t^2 = 1
t^2 = 1/4k^2
t = ±1/(2k)

So, t in terms of k is t = ±1/(2k).

b) To find the coordinates of the point of intersection of the tangent lines, we can substitute t = ±1/(2k) back into the equation f(x) = kx^2 + 3.

For t = 1/(2k):
f(1/(2k)) = k(1/(2k))^2 + 3
f(1/(2k)) = k(1/(4k^2)) + 3
f(1/(2k)) = 1/(4k) + 3

For t = -1/(2k):
f(-1/(2k)) = k(-1/(2k))^2 + 3
f(-1/(2k)) = k(1/(4k^2)) + 3
f(-1/(2k)) = 1/(4k) + 3

So, the coordinates of the point of intersection of the tangent lines are (1/(4k) + 3, 1/(2k)) and (1/(4k) + 3, -1/(2k)) in terms of k.

To find the value of t in terms of k for part (a), we need to determine when the tangent lines to the graph of f at (t, f(t)) and (-t, f(-t)) are perpendicular.

Step 1: Find the slopes of the tangent lines.
The slope of the tangent line to the graph of f at (t, f(t)) can be found by taking the derivative of f(x) and evaluating it at x = t.

f'(x) = 2kx
The slope of the tangent line at (t, f(t)) is given by f'(t) = 2kt.

Similarly, the slope of the tangent line to the graph of f at (-t, f(-t)) is given by f'(-t) = 2k(-t) = -2kt.

Step 2: Determine when the slopes are perpendicular.
For two lines with slopes m1 and m2, the lines are perpendicular if and only if the product of their slopes is -1.
So, we have:

m1 * m2 = (2kt) * (-2kt) = -4k^2t^2

Since we want the tangent lines to be perpendicular, their slope product should be -1:

-4k^2t^2 = -1

Step 3: Solve for t.

Divide both sides of the equation by -4k^2 to isolate t^2:

t^2 = 1 / (4k^2)

Finally, take the square root of both sides:

t = ±sqrt(1 / (4k^2)) = ±1 / (2k)

Therefore, the value of t in terms of k is t = ±1 / (2k).

For part (b), we need to find the coordinates of the point of intersection of the tangent lines mentioned in part (a) in terms of k.

We already found the slopes of the tangent lines:

The slope of the tangent line at (t, f(t)) is given by f'(t) = 2kt.
The slope of the tangent line at (-t, f(-t)) is given by f'(-t) = -2kt.

Now we can find the coordinates of the point of intersection using the point-slope form of the line equation.

For the tangent line at (t, f(t)):
y - f(t) = 2kt(x - t)
y = 2kt(x - t) + f(t)

For the tangent line at (-t, f(-t)):
y - f(-t) = -2kt(x + t)
y = -2kt(x + t) + f(-t)

To find the point of intersection, set these two equations equal to each other:

2kt(x - t) + f(t) = -2kt(x + t) + f(-t)

Simplify:
2ktx - 2kt^2 + f(t) = -2ktx - 2ktx + f(-t)

Combine like terms:
4ktx = -2kt^2 + f(-t) - f(t)

Divide by 4kt:
x = (-2kt^2 + f(-t) - f(t)) / (4kt)

So, the x-coordinate of the point of intersection is x = (-2kt^2 + f(-t) - f(t)) / (4kt).
To find the y-coordinate, substitute the x-coordinate into either of the tangent line equations. Let's use the equation of the tangent line at (t, f(t)):

y = 2kt(x - t) + f(t)

Substitute x = (-2kt^2 + f(-t) - f(t)) / (4kt):

y = 2kt((-2kt^2 + f(-t) - f(t)) / (4kt) - t) + f(t)

Cancel out k and t terms:

y = (-2kt^2 + f(-t) - f(t)) / 2 - 2kt^2 + f(-t) - f(t)

Simplify:

y = (-2kt^2 + f(-t) - f(t)) / 2 - 2kt^2 + f(-t) - f(t) = (-4kt^2 + 2f(-t) - 2f(t)) / 2

y = -2kt^2 + f(-t) - f(t)

Therefore, the coordinate of the point of intersection of the tangent lines mentioned in part (a) in terms of k is (x, y) = ((-2kt^2 + f(-t) - f(t)) / (4kt), -2kt^2 + f(-t) - f(t)).

To solve the problem, we will use the properties of tangent lines to a curve.

a) When two lines are perpendicular to each other, the product of their slopes is -1. Therefore, to find t, we need to find the slopes of the tangent lines at (t, f(t)) and (-t, f(-t)), and set their product equal to -1.

Step 1: Find the derivative of the function f(x).
The derivative of f(x) with respect to x is given by:
f'(x) = 2kx

Step 2: Find the slope of the tangent line at (t, f(t)).
Plugging in t into the derivative, we get:
m1 = f'(t) = 2kt

Step 3: Find the slope of the tangent line at (-t, f(-t)).
Plugging in -t into the derivative, we get:
m2 = f'(-t) = 2k(-t) = -2kt

Step 4: Set the product of the slopes equal to -1 and solve for t.
m1 * m2 = -1
(2kt) * (-2kt) = -1
4k^2t^2 = -1
t^2 = -1 / (4k^2)
t = ±√(-1 / (4k^2))
Note that since we are dealing with real numbers, t can only be real if the expression under the square root is non-negative. This means that the discriminant must be greater than or equal to 0.

Discriminant = -1 / (4k^2)
To satisfy the condition, the discriminant must be greater than or equal to zero:
-1 / (4k^2) ≥ 0
This inequality is true for all values of k except k = 0.

Therefore, the value of t in terms of k is:
t = ±√(-1 / (4k^2)), for k ≠ 0

b) To find the point of intersection of the tangent lines, we need to find the coordinates (x, y) where the tangent lines intersect.

Step 1: Find the equations of the tangent lines.
The equation of a line with slope m passing through the point (x1, y1) is given by the point-slope form:
y - y1 = m(x - x1)

Using the slope-intercept form (y = mx + c), we can rewrite this equation as:
y = mx - mx1 + y1

For the first tangent line at (t, f(t)), the equation becomes:
y = m1x - m1t + f(t)

For the second tangent line at (-t, f(-t)), the equation becomes:
y = m2x - m2(-t) + f(-t)

Step 2: Solve the system of equations to find the point of intersection.
First, let's substitute m1 and m2 with their respective values:
For the first tangent line, the equation becomes:
y = 2kt * x - 2kt * t + k(t^2) + 3

For the second tangent line, the equation becomes:
y = -2kt * x + 2kt * (-t) + k(t^2) + 3

Next, let's equate the two equations and solve for x and y:
2kt * x - 2kt * t + k(t^2) + 3 = -2kt * x + 2kt * (-t) + k(t^2) + 3

Simplifying the equation:
4kt * x = 4kt * t

Dividing both sides by 4kt:
x = t

Now, substitute this value of x back into one of the equations:
y = 2kt * x - 2kt * t + k(t^2) + 3

Using x = t:
y = 2kt * t - 2kt * t + k(t^2) + 3
y = k(t^2) + 3

Therefore, the coordinates of the point of intersection of the tangent lines are:
(x, y) = (t, k(t^2) + 3)