Question

one of the foci is (0,-1) the corresponding directrix is 3x+16=0 and e=3÷5

Answer 1

Answer :

Since 0 < e < 1, the given curve be an ellipse

Here, one focus is at S (0, - 1)

Equation of the directrix (KZ) is 3x + 16 = 0

Let, any point on the ellipse be P (x, y)

PM = the perpendicular distance of the directrix from its focus

By the definition of ellipse, we get

SP = e . PM

$\implies \sqrt{ {(x - 0)}^{2} + {(y - ( - 1))}^{2} } = \frac{3}{5} | (\frac{3x + 16}{ \sqrt{ {3}^{2} + {0}^{2} } }) | \\ \\ \implies \sqrt{ {x}^{2} + {(y + 1)}^{2} } = \frac{3}{5} |( \frac{3x + 16}{3} )| \\ \\ \implies {(5 \sqrt{ {x}^{2} + {(y + 1)}^{2} } ) \: }^{2} = {(3x + 16)}^{2} \\ \\ \implies 25( {x}^{2} + {y}^{2} + 2y + 1) = 9 {x}^{2} + 96x + 256 \\ \\ \implies 25 {x}^{2} +25 {y}^{2} + 50y + 25 - 9 {x}^{2} - 96x - 256 = 0 \\ \\ \implies 16 {x}^{2} + 25 {y}^{2} - 96x + 50y - 231 = 0$

which is the required ellipse.

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