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Exercise 5.6 · Q20

Q.The eccentricity of the ellipse (x−3)2+(y−4)2=y29(x-3)^2+(y-4)^2=\dfrac{y^2}9 is

(1) 32\dfrac{\sqrt3}2
(2) 13\dfrac13
(3) 132\dfrac1{3\sqrt2}
(4) 13\dfrac1{\sqrt3}
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Recognise the equation directly in the form SP2=e2PM2SP^2=e^2PM^2 — no standard-form conversion is needed at all.

Step 1. Match to the conic definition. (x−3)2+(y−4)2(x-3)^2+(y-4)^2 is the squared distance from (3,4)(3,4) (the focus SS); y29=(y3)2\dfrac{y^2}9=\left(\dfrac y3\right)^2, and yy (i.e. ∣y−0∣|y-0|) is exactly the perpendicular distance from (x,y)(x,y) to the line y=0y=0 (the xx-axis, the directrix). …

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