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AP EAMCET · Maths · Ellipse

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

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

Answer & Solution

Correct Answer

(D) \(\frac{3}{\sqrt{2}}\)

Step-by-step Solution

Detailed explanation

Let \(A=(0,3)\) Given the ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\) So focus of the ellipse \(F=\left(\sqrt{5^2-4^2}, 0\right)\) \(\Rightarrow \mathrm{F}=(3,0), \mathrm{P}\left(\frac{0+3}{2}, \frac{3+0}{2}\right)=\left(\frac{3}{2}, \frac{3}{2}\right)\) So required distance…