9514 1404 393
Answer:
- tangent point: (2, 4)
- normal: y = -x +6
Step-by-step explanation:
The derivative of P with respect to t is ...
P' = (x', y') = (4t, 4)
so the slope of the curve for some value of t is ...
dy/dx = y'/x' = 4/(4t) = 1/t
The line we want to be a tangent is y = x +2, which has a slope of 1. The curve will have a slope of 1 where ...
1/t = 1 ⇒ t = 1
At t=1, the point P is (2·1², 4·1) = (2, 4). For x=2, the point on the desired tangent is y = x +2 = 2 +2 = 4, or (x, y) = (2, 4).
The curve and the given line both have a slope of 1 at the point (2, 4), so that is the tangent point.
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The normal to the curve at (2, 4) will have a slope that is the opposite reciprocal of the slope of the tangent: -1/1 = -1. Then point-slope form of the equation of the normal line is ...
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
y -4 = -1(x -2) . . . . . . line with slope -1 through point (2, 4)
y = -x +6 . . . . . . . . equation of the normal line in slope-intercept form
