Question

find the derivative of y²=8(x-6)
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Answer 1

Answer:

This parabola has a vertex at (0,0) and opens to the right along the positive x axis

It is also of the form y2 = 4ax, dx/dt = 4a

y2 = 8x

However we are given a point that lies inside the parabola (4, 2) instead of on the parabola

Graph the parabola, draw a line that passes through (4,2) and the shortest distance to one point on the parabola, use these two points and calculate the slope of the line (Normal Line).

I used (4,2) and (2,4), got a slope of negative 1 and line and y intercept at (0,6)

y = -x + 6

The line perpendicular to this line that passes through (2,4) will be a tangent to your parabola.

You can graph y2 = 8x at Desmos.com

Also

Once you get a point on the parabola if you are familiar with derivatives you can use the derivative approach above.

In either case I got y = x + 2 for a tangent.

There is more than one tangent to the parabola

Give it try.

Answer 2

Answer:

$y' = 4/y$

Step-by-step explanation:

=> $\frac{d[y^2]}{dx} = \frac{d[8(x-6)]}{dx}$

$=> 2y.\frac{d[y]}{dx} = 8 (\frac{d[x]}{dx} + \frac{d[-6]}{dx})$

=> $2yy' = 8(1+0)$

=> $2yy' = 8$

=> $y' = \frac{4}{y}$