Consider the particle travelling 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

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

$(0, \frac{- 25}{3})$

(0, 9)

(0, 7/3)

answer is A.

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Detailed Solution

Equation of an ellipse is $\frac{x^{2}}{100} + \frac{y^{2}}{25} = 1$
$E q u a t i o n o f \tan g e n t a t P (- 8, 3) i s \frac{x (- 8)}{100} + \frac{y (3)}{25} = 1 \Rightarrow 2 x - 3 y + 25 = 0 f o r y - a x i s p u t x = 0 t h e n y = \frac{25}{3} S o r e q u i r e d p o i n t = (0, \frac{25}{3})$