Question

PSQ is a focal chord of the ellipse 4x² + 9y² = 36 such that SP = 4. If S' is the another focus, write the value of S'Q. 

Answer 1

\[\text{ The given equation of the ellipse is } \]
\[4 x^2 + 9 y^2 = 36 . . . (i)\]
\[ \Rightarrow \frac{x^2}{9} + \frac{y^2}{4} = 1\]
\[\text{ This is of the form } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \text{ where } a^2 = 9 \text{ and } b^2 = 4 i . e a = 3 \text{ and } b = 2\]
\[\text{ Clearly } a > b, \text{ therefore the major axis and minor axis of the ellipse are along } \]
\[\text{ x axis and y axis respectively } . \]
\[\text{ Let , e be the ecentricity of the ellipse . Then }, \]
\[ \therefore e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{4}{9}}\]
\[ = \frac{\sqrt{5}}{3}\]
\[\text{ Therefore, coordinates of focus at S i . e } \left( ae, 0 \right) = \left( \sqrt{5}, 0 \right)\]
\[\text{ & coordinates of focus at } S' = \left( - ae, 0 \right) = \left( - \sqrt{5}, 0 \right)\]
\[\text{ It is given that PSQ is a focal chord } \]
\[\text{ As we know that } , \]
\[SP + S'P = 2a\]
\[ \Rightarrow 4 + S'P = 6 \left[ \because SP = 4 \right]\]
\[ \Rightarrow S'P = 2\]
\[\text{ Let coordinates of P be } \left( m, n \right)\]
\[\text{ As S'P } = 2\]
\[ \therefore \sqrt{\left( n - 0 \right)^2 + \left( m + \sqrt{5} \right)^2} = 2\]
\[ \Rightarrow n^2 + m^2 + 5 + 2\sqrt{5}m = 4 . . . (ii)\]
\[\text{ & SP } = 4\]
\[ \therefore \sqrt{\left( n - 0 \right)^2 + \left( m - \sqrt{5} \right)^2} = 4\]
\[ \Rightarrow n^2 + m^2 + 5 - 2\sqrt{5}m = 16 . . . (iii)\]
\[\text{ Subtracting eq } (iii) \text{ from eq } (ii) , \text{ we get } \]
\[4\sqrt{5}m = - 12\]
\[ \Rightarrow m = \frac{- 3}{\sqrt{5}}\]
\[\text{ & we get n } = \frac{4}{\sqrt{5}}\]
\[ \therefore \text{ Coordinates of P are } \left( \frac{- 3}{\sqrt{5}}, \frac{4}{\sqrt{5}} \right) \text{ and coordinates of S are } \left( \sqrt{5}, 0 \right)\]
\[ \therefore \text{ Equation of the line segment PS which is extended to PQ is given by }\]
\[ x + 2y = \sqrt{5} . . . (iv)\]
\[\text{ Solving eq (i) & (iv) we get } , \]
\[x = \frac{- 3}{\sqrt{5}} \text{ and } y = \frac{4}{\sqrt{5}} \text{ which are the coordinates of P }\]
\[\text{ & } x = \frac{66\sqrt{5}}{50} \text{ and } y = \frac{- 8\sqrt{5}}{50} \text{ which would be the coordinates of Q } . \]
\[ \therefore S'Q = \sqrt{\left( \frac{- 8\sqrt{5}}{50} - 0 \right)^2 + \left( \frac{66\sqrt{5}}{50} + \sqrt{5} \right)^2}\]
\[ = \frac{26}{5}\]