Question:
The product of perpendiculars from the two foci of the ellipse
\[
\frac{x^2}{9} + \frac{y^2}{25} = 1
\]
on the tangent at any point on the ellipse is:
The product of perpendiculars from the two foci of the ellipse
\[
\frac{x^2}{9} + \frac{y^2}{25} = 1
\]
on the tangent at any point on the ellipse is:
Show Hint
For standard ellipses, the product of perpendicular distances from the foci to any tangent is always equal to \( b^2 \).
Updated On: Mar 24, 2025
- \( 6 \)
- \( 7 \)
- \( 8 \)
- \( 9 \)
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The Correct Option is D
Solution and Explanation
Step 1: Using the focal property of an ellipse
For an ellipse, the product of the perpendicular distances from the foci to a tangent at any point is given by:
\[
b^2.
\]
Here, \( a^2 = 25 \), \( b^2 = 9 \), so the product of the perpendicular distances is:
\[
9.
\]
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