9. 2x2 + 2y2 – x = 0

9. 2x2 + 2y2= x2(x2 + y2) = xx2 + y2 = x2x2 – x2 + y2= 0x2 – 2 (x4) + y2 = 0x2 – 2(x4)+(14)2−(14)2+y2=0x2 – 2(14x)+(14)2+y2=(14)2Using,(a – b)2 = a2 + b2 – 2ab,We get(x−14)2+(y−0)2=(14)2Comparing with the equation of a circle (x – h)2 + (y – k)2 = r2We get,h = 14 , k = 0, r = 14?Centre = (h, k) = (14,0)Radian = r = 14

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Conic Sections Ncert Solutions Maths class 11th Maths Class 11th Conic Sections

9. 2x² + 2y² – x = 0

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

9. 2x² + 2y²= x

2(x² + y²) = x

x² + y² = $\frac{x}{2}$

x² – $\frac{x}{2}$ + y²= 0

x² – 2 $(\frac{x}{4})$ + y² = 0

x² – $2 (\frac{x}{4}) + {(\frac{1}{4})}^{2} - {(\frac{1}{4})}^{2} + y^{2} = 0$

x² – $2 (\frac{1}{4} x) + {(\frac{1}{4})}^{2} + y^{2} = {(\frac{1}{4})}^{2}$

Using,

(a – b)² = a² + b² – 2ab,

We get

${(x - \frac{1}{4})}^{2} + {(y - 0)}^{2} = {(\frac{1}{4})}^{2}$

Comparing with the equation of a circle (x – h)² + (y – k)² = r²

We get,

h = $\frac{1}{4}$ , k = 0, r = $\frac{1}{4}$

?Centre = (h, k) = $(\frac{1}{4}, 0)$

Radian = r = $\frac{1}{4}$

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Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

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