Question

If a focal chord of the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) meets its minor axis at the point \((0, 3)\), then the perpendicular distance from the centre of the ellipse to this focal chord is:

• A. \(5\)
• B. \(\frac{2}{\sqrt{5}}\)
• C. \(1\)
• D. \(\frac{3}{\sqrt{2}}\)

Hint:

For perpendicular from center to focal chord, use point-to-line distance formula and symmetry of ellipse.

Answer 1

The Correct Option is D

Focal chord of ellipse intersects minor axis at \((0, 3)\). Center of ellipse = origin. Chord passes through (0, 3) and is symmetric about major axis. The focal chord is a line segment joining two points on the ellipse and passing through one focus. Let the equation of the focal chord be \(y = mx + c\) such that it passes through (0,3), so \(c = 3\) The perpendicular distance from the center (0,0) to this line is: \[ \text{Distance} = \frac{|3|}{\sqrt{m^2 + 1}} \] We must find the correct \(m\) based on it being a focal chord of this ellipse. Instead, recognize that since (0,3) lies on focal chord and minor axis, and the chord is symmetric, the perpendicular from center to chord is perpendicular to line passing through (0,3), i.e., the perpendicular distance from (0,0) to line passing through focus and (0,3). Using geometry, this distance turns out to be \(\frac{3}{\sqrt{2}}\)