Question

Consider the particle traveling clockwise on the elliptical path $\frac{x^{2}}{100}$+ $\frac{y^{2}}{25}$= 1. The particle leaves the orbit at the point

(–8, 3) and travels in a straight line tangent to the ellipse.

At what point will the particle cross the y-axis:

• A. $\left( 0,\frac{25}{3} \right)$
• B. $\left( 0, - \frac{25}{3} \right)$
• C. (0, 9)
• D. $\left( 0,\frac{7}{3} \right)$

Answer 1

Answer: (A)

$\left( 0,\frac{25}{3} \right)$

Answer 2

General points on ellipse (10 cos q, 5 sin q)

tangent $\frac{x\cos\theta}{10}$+ $\frac{y\sin\theta}{5}$= 1

through (–8, 3)

– $\frac{8}{10}$cos q + $\frac{3}{5}$sin q = 1

Ž sin q. $\frac{3}{5}$– cos q. $\frac{4}{5}$= 1

cos f = $\frac{3}{5}$, sin f = $\frac{4}{5}$ y- coordinate

sin (q – f) = 1 = $\frac{5}{\sin\theta}$

q = $\frac{\pi}{2}$+ f = $\frac{5}{\cos\varphi}$= $\frac{25}{3}$