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NCERT Exemplar · Q30

Q.Find the equation of the set of all points whose distance from (0,4)(0, 4) are 23\dfrac{2}{3} of their distance from the line y=9y = 9.

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Setting distance from (0,4)(0,4) equal to 23\tfrac{2}{3} of the distance from y=9y=9 and simplifying gives the ellipse 9x2+5y2=1809x^2 + 5y^2 = 180, i.e. x220+y236=1\dfrac{x^2}{20} + \dfrac{y^2}{36} = 1.

Let P(x,y)P(x, y) be any point in the set. The condition is

x2+(y−4)2=23 ∣y−9∣\sqrt{x^2 + (y-4)^2} = \frac{2}{3}\,|y - 9|

1. Square both sides.

x2+(y−4)2=49(y−9)2x^2 + (y-4)^2 = \frac{4}{9}(y-9)^2

2. Multiply by 99 and expand.

9x2+9(y2−8y+16)=4(y2−18y+81)9x^2 + 9(y^2 - 8y + 16) = 4(y^2 - 18y + 81)

9x2+9y2−72y+144=4y2−72y+3249x^2 + 9y^2 - 72y + 144 = 4y^2 - 72y + 324

3. Simplify.

9x2+5y2−180=0  ⇒  9x2+5y2=1809x^2 + 5y^2 - 180 = 0 \;\Rightarrow\; 9x^2 + 5y^2 = 180

4. Standard form. Dividing by 180180: …

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