Maximum distance between two points lying on the curve $4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0$ is

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Solution

Given curve is,
$4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0$
$\Rightarrow 4 (x^{2} + 2 x) + 9 (y^{2} - 4 y) = - 4$
$\Rightarrow 4 (x^{2} + 2 x + 1) + 9 (y^{2} - 4 y + 4) = - 4 + 4 + 36 = 36$
$\Rightarrow \frac{(x + 1)^{2}}{9} + \frac{(y - 2)^{2}}{4} = 1$
Clearly this is an ellipse,So maximum distance between two point lying is just the length of the transverse axis which is $= 2 a = 6$

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Q. Latus rectum of ellipse

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

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4 x^{2} + 9 y^{2} + 8 x + 36 y + 4 = 0

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Q. Find the length of latus rectum of the ellipse

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If C is the center and A,B are two points on the conic ${4 x}^{2}$ + ${9 y}^{2}$ -8 $x$ -36y +4=0
$∠$ ACB = $\frac{π}{2}$ , then ${C A}^{- 2}$ + ${C B}^{- 2}$ + $\frac{23}{36}$ =

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Maximum distance between two points lying on the curve 4x2+9y2+8x−36y+4=0 is

Given curve is, 4x2+9y2+8x−36y+4=0⇒4(x2+2x)+9(y2−4y)=−4⇒4(x2+2x+1)+9(y2−4y+4)=−4+4+36=36⇒(x+1)29+(y−2)24=1 Clearly this is an ellipse,So maximum distance between two point lying is just the length of the transverse axis which is =2a=6

Maximum distance between two points lying on the curve $4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0$ is