The points of intersection of the line $2 x + y = 3$ and the ellipse $4 x^{2} + y^{2} = 5$ are:

(\frac{1}{2}, 2), (1, 1)

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(\frac{1}{2}, 2), (- 1, 1)

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(\frac{- 1}{2}, 2), (- 1, 1)

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(\frac{- 1}{2}, 2), (1, 1)

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Solution

The correct option is A $(\frac{1}{2}, 2), (1, 1)$
$y = 3 - 2 x$
$4 x^{2} + (3 - 2 x)^{2} = 5$
$4 x^{2} + 4 x^{2} + 9 - 12 x = 5$
$8 x^{2} - 12 x + 4 = 0$
$2 x^{2} - 3 x + 1 = 0$
$2 x^{2} - 2 x - x + 1 = 0$
$2 x (x - 1) - 1 (x - 1) = 0$
$x = 1, \frac{1}{2}$
$\Rightarrow y = 1, 2$
$∴$ points are $(1, 1), (\frac{1}{2}, 2)$

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The points of intersection of the line 2x+y=3 and the ellipse 4x2+y2=5 are:

The correct option is A (12,2),(1,1) y=3−2x4x2+(3−2x)2=54x2+4x2+9−12x=58x2−12x+4=02x2−3x+1=02x2−2x−x+1=02x(x−1)−1(x−1)=0x=1,12⇒y=1,2∴ points are (1,1),(12,2)

The points of intersection of the line $2 x + y = 3$ and the ellipse $4 x^{2} + y^{2} = 5$ are: