The lines 3x + 4y - 5 = 0 cuts the curve 2x2 + 3y2 = 5 at A and B. If ‘O’ is the origin, then ∠AOB=

2x2 + 3y2 = 5 → (1)3x - 4y - 5 = 03x+4y5=1Homogensing (1) with help of (2)x2 - y2 - 24xya + b = 0θ=π2

The lines 3x + 4y - 5 = 0 cuts the curve 2x2 + 3y2 = 5 at A and B. If ‘O’ is the origin, then ∠AOB=

2x2 + 3y2 = 5 → (1)3x - 4y - 5 = 03x+4y5=1Homogensing (1) with help of (2)x2 - y2 - 24xya + b = 0θ=π2

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The lines 3x + 4y - 5 = 0 cuts the curve 2x² + 3y² = 5 at A and B. If ‘O’ is the origin, then $∠ AOB =$

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$π$ /6

$π$ /3

$π$ /2

$π$ /8

answer is C.

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Detailed Solution

2x² + 3y² = 5 → (1)
3x - 4y - 5 = 0
$\frac{3 x + 4 y}{5} = 1$
Homogensing (1) with help of (2)
x² - y² - 24xy
a + b = 0
$θ = \frac{π}{2}$

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The lines 3x + 4y - 5 = 0 cuts the curve 2x2 + 3y2 = 5 at A and B. If ‘O’ is the origin, then ∠AOB=

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The lines 3x + 4y - 5 = 0 cuts the curve 2x² + 3y² = 5 at A and B. If ‘O’ is the origin, then $∠ AOB =$