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
- A \(5\)
- B \(\frac{2}{\sqrt{5}}\)
- C \(1\)
- D \(\frac{3}{\sqrt{2}}\)
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…
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