Multiple choice

$\displaystyle C_1$ is a circle on the major axis and $\displaystyle C_2$ is a circle on the minor axis of the ellipse $\displaystyle P : \frac {x^2}{25} + \frac {y^2}{16} = 1$ as diameters. If $l$ is the length of the tangent from any point on the circle $\displaystyle C_1$ to the ellipse $P$, then

  1. $\displaystyle l < \dfrac23$
  2. $\displaystyle l = 3$
  3. $\displaystyle l \leq \dfrac34$
  4. $\displaystyle l = \dfrac34$
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C Correct answer
Explanation

The circle C1 has the major axis (length 10) as its diameter, so its radius is 5. The ellipse P is x^2/25 + y^2/16 = 1. The length of a tangent from a point (x1, y1) to an ellipse is given by the square root of the power of the point with respect to the ellipse. For points on the circle x^2 + y^2 = 25, the calculation shows the length l satisfies l <= 3/4.

AI explanation

For the ellipse x^2/25 + y^2/16 = 1, the major axis has endpoints at (5, 0) and (-5, 0), making the circle C1 on the major axis x^2 + y^2 = 25. The length of the tangent from a point (x1, y1) on this circle to the ellipse is given by the formula l = sqrt(S1), where S1 = x1^2/25 + y1^2/16 - 1. Substituting x1^2 = 25 - y1^2 into the expression for S1 yields (25 - y1^2)/25 + y1^2/16 - 1, which simplifies to 9y1^2/400. Since the point lies on the circle C1, the maximum value of y1^2 is 25. Therefore, the maximum length of the tangent l is sqrt(9 * 25 / 400) = 15/20 = 3/4, establishing that the length l is less than or equal to 3/4.

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