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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the coordinate of double ordinate for the ellipse $\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}$$=1$

$\begin{array}{1 1}(a)\;k=\large\frac{b}{a}\normalsize (a^2-h^2),k'=-\large\frac{b}{a}\normalsize (a^2-h^2)\\(b)\;k=\large\frac{b^2}{a}\normalsize (a^2-h^2),k'=-\large\frac{b^2}{a}\normalsize (a^2-2h^2)\\(c)\;k=\large\frac{b}{2a}\normalsize (a^2-h^2),k'=-\large\frac{b}{2a}\normalsize (a^2+h^2)\\(d)\;k=\large\frac{2b}{a}\normalsize (a^2-h^2),k'=-\large\frac{2b}{a}\normalsize (a^2-h^2)\end{array}$

1 Answer

Let $(h,k)$ & $(h,k')$ be the double ordinate which also will satisfy the ellipse also hence
$\large\frac{h^2}{a^2}+\frac{k^2}{b^2}$$=1$
$\large\frac{k^2}{b^2}=1-\large\frac{h^2}{a^2}$
$k^2=b^2(1-\large\frac{h^2}{a^2})$
$k=\large\frac{b}{a}\normalsize (a^2-h^2)$[for first quadrant)
$k'=\large\frac{b}{a}\normalsize (a^2-h^2)$[for fourth quadrant)
Hence (a) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
 

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