Question

Consider the line $L(t) = (4 + 2t, 18 + 4t)$. Then $L$ intersects: 1. The $x$-axis at the point $(0, 10)$ when $t = -2$ 2. The $y$-axis at the point $(-5, 0)$ when $t = -9/2$ 3. The parabola $y = x^2$ at the points $(1 + 2\sqrt{11}, 12 + 4\sqrt{11})$ and $(1 - 2\sqrt{11}, 12 - 4\sqrt{11})$ when $t = -3/2 + \sqrt{11}$ and $-3/2 - \sqrt{11}$

          Consider the line $L(t) = (4 + 2t, 18 + 4t)$. Then $L$ intersects:
1. The $x$-axis at the point $(0, 10)$ when $t = -2$
2. The $y$-axis at the point $(-5, 0)$ when $t = -9/2$
3. The parabola $y = x^2$ at the points $(1 + 2\sqrt{11}, 12 + 4\sqrt{11})$ and $(1 - 2\sqrt{11}, 12 - 4\sqrt{11})$ when $t = -3/2 + \sqrt{11}$ and $-3/2 - \sqrt{11}$

Consider the line L(t) = (4 + 2t, 18 + 4t). Then L intersects:
1. The x-axis at the point (0, 10) when t = -2
2. The y-axis at the point (-5, 0) when t = -9/2
3. The parabola y = x^2 at the points (1 + 2√(11), 12 + 4√(11)) and (1 - 2√(11), 12 - 4√(11)) when t = -3/2 + √(11) and -3/2 - √(11)

Added by James J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert James Kiss

09/18/2023

Step 1

Step 2: To find the point where the line intersects the y-axis, we set x=0 and solve for t: 0 = 4 + 2t18 + 4t 0 = 4 + 36t + 4t 0 = 4 + 40t -4 = 40t t = -4/40 t = -1/10 Substitute t = -1/10 into the equation It=4+2t18+4t to find the y-coordinate: y = 4 + Show more…

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Consider the line It=4+2t18+4t. Then Lintersect: 1. The x-axis at the point 0,10 when t= 2. The y-axis at the point -5,0 when t= -9/2 3. The parabola y=at the points 1+2sqrt(11),12+4sqrt(11) and (1-2sqrt(11),12-4sqrt(11)) when t= -3/2+sqrt(11) -3/2-sqrt(11)