Question:
Find the points of intersection of the given surface
\[
\frac{x^2}{81} + \frac{y^2}{36} + \frac{z^2}{4} = 1 \text{ and the straight line } \frac{x - 3}{3} = \frac{y - 4}{-6} = \frac{z + 2}{4}
\]
Find the points of intersection of the given surface
\[
\frac{x^2}{81} + \frac{y^2}{36} + \frac{z^2}{4} = 1 \text{ and the straight line } \frac{x - 3}{3} = \frac{y - 4}{-6} = \frac{z + 2}{4}
\]
Updated On: Mar 30, 2025
- (3, 4, -1)
- (6, -2, 2)
- (3, 4, -2) and (6, -2, 2)
- (-3, 4, -2) and (6, -2, 2)
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The Correct Option is C
Solution and Explanation
Parametrize line with parameter \(t\): \[ x = 3t + 3,\quad y = -6t + 4,\quad z = 4t - 2 \] Substitute into surface: \[ \frac{(3t + 3)^2}{81} + \frac{(-6t + 4)^2}{36} + \frac{(4t - 2)^2}{4} = 1 \] Solve this equation to find values of \(t\), then plug back into line equations to get coordinates. Final points are: (3, 4, -2) and (6, -2, 2).
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