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: 

• A. \( 6 \)
• B. \( 7 \)
• C. \( 8 \)
• D. \( 9 \)

Hint:

For standard ellipses, the product of perpendicular distances from the foci to any tangent is always equal to \( b^2 \).

Answer 1

The Correct Option is D

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: \[ 9. \]

Answer 2

Given the ellipse:
\[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \] and the product of perpendiculars from its two foci on the tangent at any point on the ellipse is to be found.

Step 1: Identify ellipse parameters:
\[ a^2 = 25, \quad b^2 = 9 \] Since \(a^2 > b^2\), the major axis is along the y-axis.
Foci are located at:
\[ (0, \pm c), \quad c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]

Step 2: Equation of tangent at point \((x_0, y_0)\) on ellipse:
\[ \frac{x x_0}{9} + \frac{y y_0}{25} = 1 \]

Step 3: Perpendicular distance \( d \) from point \( (x_1, y_1) \) to line \( Ax + By + C = 0 \) is:
\[ d = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}} \] Rewrite tangent line:
\[ \frac{x x_0}{9} + \frac{y y_0}{25} - 1 = 0 \] Multiply through by \(225\) (LCM of 9 and 25) for simplicity:
\[ 25 x x_0 + 9 y y_0 - 225 = 0 \] So: \[ A = 25 x_0, \quad B = 9 y_0, \quad C = -225 \]

Step 4: Calculate perpendicular distances from foci \( F_1(0,4) \) and \( F_2(0,-4) \):
\[ d_1 = \frac{|25 x_0 \times 0 + 9 y_0 \times 4 - 225|}{\sqrt{(25 x_0)^2 + (9 y_0)^2}} = \frac{|36 y_0 - 225|}{\sqrt{625 x_0^2 + 81 y_0^2}} \] \[ d_2 = \frac{|25 x_0 \times 0 + 9 y_0 \times (-4) - 225|}{\sqrt{625 x_0^2 + 81 y_0^2}} = \frac{|-36 y_0 - 225|}{\sqrt{625 x_0^2 + 81 y_0^2}} = \frac{|36 y_0 + 225|}{\sqrt{625 x_0^2 + 81 y_0^2}} \]

Step 5: Product of perpendiculars:
\[ d_1 d_2 = \frac{|36 y_0 - 225| \times |36 y_0 + 225|}{625 x_0^2 + 81 y_0^2} = \frac{|(36 y_0)^2 - (225)^2|}{625 x_0^2 + 81 y_0^2} \] \[ = \frac{|1296 y_0^2 - 50625|}{625 x_0^2 + 81 y_0^2} \]

Step 6: Using the ellipse equation:
\[ \frac{x_0^2}{9} + \frac{y_0^2}{25} = 1 \implies 25 x_0^2 + 9 y_0^2 = 225 \] Multiply both sides by 25:
\[ 625 x_0^2 + 225 y_0^2 = 5625 \] Substitute \( 625 x_0^2 = 5625 - 225 y_0^2 \) into denominator:
\[ 625 x_0^2 + 81 y_0^2 = (5625 - 225 y_0^2) + 81 y_0^2 = 5625 - 144 y_0^2 \]

Step 7: Substitute in numerator and denominator:
\[ d_1 d_2 = \frac{|1296 y_0^2 - 50625|}{5625 - 144 y_0^2} \]

Step 8: Check value at point on ellipse \( y_0^2 = 25 \) (since \( b^2 = 9 \) and \( a^2 = 25 \), choose a point that satisfies the ellipse). For example, at \( y_0 = 5 \):
\[ d_1 d_2 = \frac{|1296 \times 25 - 50625|}{5625 - 144 \times 25} = \frac{|32400 - 50625|}{5625 - 3600} = \frac{18225}{2025} = 9 \]

Therefore, the product of perpendiculars from the two foci on the tangent at any point on the ellipse is constant and equal to:
\[ \boxed{9} \]