Consider the function F(x)=k(x^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 term of k.

b) Find the slopes of tangent lines mentioned in part (a).

c) Find the coordinates of the point of intersection of the tangent lines mentioned in part (a).

F(x)=k(x^2)+3

= kx^2 + 3
F'(x) = 2kx
at( t, f(t)) slope of tangent is 2kt
at (-t, f(-t), slope of tangent is -2kt
but they are perpendicular, so
(2kt)(-2kt) = -1
4k^2t^2= 1
k^2 t^2 = 1/4
kt = ± 1/2
t = ± 1/ (2k)
when t = 1/(2k)
so at( t, f(t)) slope of tangent is 2kt = 2k(1/(2k)) = 1
at (-t, f(-t), slope of tangent is -2kt = -2k(1/(2k)) = -1

c) how about you try it ?

a) Well, if the tangent lines at (t,F(t)) and (-t,F(-t)) are perpendicular, then their slopes must be negative reciprocals of each other. Let's find the slopes first!

b) The slope of the tangent line at (t,F(t)) can be found by taking the derivative of F(x) and evaluating it at x=t. So, let's do that: F'(x) = 2kt, and at x=t, the slope is 2kt.

Similarly, the slope of the tangent line at (-t,F(-t)) is 2kt, since the derivative of F(x) is the same regardless of the sign of x.

c) Now, since the slopes of the tangent lines are negative reciprocals of each other, we have 2kt = -1/(2kt). Let's solve this equation for t in terms of k:

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

Hmm, this is an imaginary solution. It seems like there is no real value of t that satisfies the condition for the tangent lines to be perpendicular. So, the answer is unfortunately "No real solution, sorry!"

But hey, chin up! At least we had some fun with imaginary numbers, right?

To solve this problem, we'll need to find the derivative of the function F(x), and then use that information to find the tangent lines and their slopes.

a) To find t in terms of k when the tangent lines are perpendicular, we need to find the derivative of F(x) and set their slopes equal to the negative reciprocals of each other.

The derivative of F(x) is given by:

F'(x) = 2kx

Let's find the slopes of the tangent lines in terms of t:

The slope of the tangent line at (t, F(t)) is given by F'(t), so it is 2kt.

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

Since the slopes of perpendicular lines are negative reciprocals of each other, we have:

2kt = -1/(-2kt)

Cross multiplying gives:

4k^2t^2 = 1

Dividing both sides by 4k^2 gives:

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

Taking the square root of both sides gives:

t = ±1/(2k)

Therefore, t is equal to ±1/(2k).

b) We already found the slopes of the tangent lines in part (a):

The slope of the tangent line at (t, F(t)) is 2kt.

The slope of the tangent line at (-t, F(-t)) is -2kt.

c) To find the coordinates of the point of intersection of these tangent lines, we need to find the y-coordinate by plugging in the x-coordinate into the original function F(x).

Let's find the y-coordinate when x = t:

F(t) = k(t^2) + 3

And the y-coordinate when x = -t:

F(-t) = k((-t)^2) + 3

Simplifying these expressions:

F(t) = kt^2 + 3

F(-t) = kt^2 + 3

Since they both have the same y-coordinate, they intersect at the same point.

Therefore, the coordinates of the point of intersection of the tangent lines are (t, kt^2 + 3).

To solve this problem, we need to find the value of t in terms of k that makes the tangent lines perpendicular, find the slopes of those tangent lines, and finally find the coordinates of their point of intersection.

a) To determine when two lines are perpendicular, we can check if the product of their slopes is equal to -1. So, let's find the slopes of the tangent lines at (t, F(t)) and (-t, F(-t)) first.

To find the slope of a tangent line, we need to find the derivative of the function F(x). Taking the derivative of F(x) = kx^2 + 3 gives us:

F'(x) = 2kx

Now, let's find the slope of the tangent line at (t, F(t)):

F'(t) = 2kt

The slope of a line perpendicular to this tangent line will be the negative reciprocal of its slope:

Slope of perpendicular line = -1 / (2kt)

Similarly, let's find the slope of the tangent line at (-t, F(-t)):

F'(-t) = 2k(-t) = -2kt

Again, the slope of a line perpendicular to this tangent line will be the negative reciprocal:

Slope of perpendicular line = -1 / (-2kt) = 1 / (2kt)

Now, we can set up the equation for perpendicular lines:

Product of slopes = (slope of first line) * (slope of second line) = -1

(-1 / (2kt)) * (1 / (2kt)) = -1

Simplifying the equation:

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

Multiplying both sides by 4k^2t^2:

1 = -4k^2t^2

Dividing both sides by -4k^2:

-1/4k^2 = t^2

Take the square root of both sides:

t = ± sqrt(-1/4k^2) = ± i * (1 / (2k))

Therefore, t = ± i / (2k).

b) Let's find the slopes of the tangent lines mentioned in part (a). We already determined the slope of the tangent line at (t, F(t)) to be 2kt. The slope of the tangent line at (-t, F(-t)) is -2kt.

c) Finally, let's find the coordinates of the point of intersection of the tangent lines mentioned in part (a).

To find this point, we can substitute the x-coordinate (t) into the original function F(x).

F(t) = k(t^2) + 3

And substitute the x-coordinate (-t) into the original function F(x).

F(-t) = k((-t)^2) + 3 = k(t^2) + 3

Since both F(t) and F(-t) are equal, the y-coordinates of the point of intersection are the same. We can set the two equations equal to each other:

k(t^2) + 3 = k(t^2) + 3

The constants cancel out:

0 = 0

This equation is always true regardless of the value of t. Therefore, the coordinates of the point of intersection are the same for all values of t.