Hello, Confuzzled!

Can't read the first one . . .

Eliminate the parameter:b) The curveChas parametric equations: .x = 4·cos2t and y = 3·sin t

for -π/2 < t < π/2.

Ais the point (2,3/2) and lies onC.

The normal toCatAis: .6y - 16x + 23 .= .0.

The normal atAcutsCagain at pointB.

Find the y-coordinate of pointB.

. . x .= .4·cos2t .= .4(1 - 2·sin²t) .[1]

. . y .= .3·sin t . → . sin t .= .y/3[2]

Substitute [2] into [1]: .x .= .4(1 - 2[y²/9])

. . and we get: .x .= .4 - 8y²/9 .[3]. . . a parabola

The equation of the normal is: .x .= .3y/8 + 23/16 .[4]

Equate [3] and [4]: .4 - 8y²/9 .= .3y/8 + 23/16

. . which simplifies to: .128y² + 54y - 369 .= .0

. . which factors: .(2y - 3)(64y + 123) .= .0 .**

. . and has roots: .y .= .3/2, -123/64

So the normal atAcuts the parabola at: .y = -123/64

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Okay, okay! . . . I didn'treallyfactor it.

I used the Quadratic Formula and was pleasantly surprised

. . to find that the roots were rational.

From the roots I constructed the factors,

. . then created theillusionthat I'm a genius.