Question

find the equations of the tangents to the curve x = 6t2 2, y = 4t3 2 that pass through the point (8, 6). y = (smaller slope) y = (larger slope)

Answer 1

To find the equations of the tangents, differentiate the given equations, find the slopes at the given point, and solve for the y-intercepts.

Step-by-step explanation:

To find the equations of the tangents to the curve, we need to find the slope of the tangent lines at the given point.

First, differentiate the given equations with respect to t.

For x = 6t^2 -2, the derivative is dx/dt = 12t.

For y = 4t^3 -2, the derivative is dy/dt = 12t^2.

Substitute the value of t = 25 into dx/dt and dy/dt to find the slopes at that point. Substitute the value of x = 8 and y = 6 into the equations, and solve for the y-intercepts.

The equation of the tangent line with the smaller slope is y = 156x - 1248, and the equation of the tangent line with the larger slope is y = 155.82x - 1245.92.

Answer 2

To find the equations of the tangents to the given curve, we need to find the slope of the tangents that pass through the point (8, 6). The slope can be found by taking the derivative of y with respect to x. Using the derivative and the given point, we can find the equations of the tangents. The equations of the tangents are y = 2x - 10 (smaller slope) and y = 2x - 14 (larger slope).

Step-by-step explanation:

To find the equations of the tangents to the curve, we need to find the slope of the tangents that pass through the point (8, 6). The curve is represented by the equations x = 6t^2 + 2 and y = 4t^3 - 2. We can find the slope of the tangents by taking the derivative of y with respect to x.

We have dx/dt = 12t and dy/dt = 12t^2. To find the slope, we calculate dy/dx = (dy/dt)/(dx/dt).

Substituting the values of dx/dt and dy/dt, we get dy/dx = (12t^2)/(12t) = t.

Now, we can find the value of t when the curve passes through the point (8, 6). We substitute x = 8 and y = 6 into the equations x = 6t^2 + 2 and y = 4t^3 - 2 to get two equations in terms of t. Solving these equations, we find that t = 2.

Finally, we substitute t = 2 into the equation dy/dx = t to find the slope of the tangent at the point (8, 6). The slope is 2.

So, the equations of the tangents are y = mx + c, where m is the slope and c is the y-intercept. Since the tangents pass through the point (8, 6), we can substitute x = 8 and y = 6 into the equations to find c. The equations of the tangents are:

y = 2x - 10 (smaller slope)

y = 2x - 14 (larger slope)