Question

Help with 5 please( question tuped below too).

Prove that y=x+2 is a tangent to the locous P(2t²,4t).

find the point of contact of this tangent to this locous and hence find the equation of normal at the point.


thanks
​

Answer 1

9514 1404 393

Answer:

• tangent point: (2, 4)
• normal: y = -x +6

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.

__

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