Question

Q.1
The line y = 2x-8 cuts the curve 2x2+ y2 - 5xy + 32 = 0) at the points A and B.
Find the length of the line AB
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Answer 1

Step-by-step explanation:

The given line is $y=2x-8$

The given curve is $2 {x}^{2} + {y}^{2} - 5xy + 32 = 0$

Now,

$2 {x}^{2} + (2x - 8)^{2} - 5x(2x - 8) + 32 = 0$

$\implies2 {x}^{2} + 4x^{2} - 32x+ 64 - 10 {x}^{2} + 40x + 32 = 0$

$\implies - 4 {x}^{2} + 8x+ 96 = 0$

$\implies {x}^{2} - 2x - 24 = 0$

$\implies {x}^{2} - 6x + 4x - 24 = 0$

$\implies x(x - 6) + 4(x - 6) = 0$

$\implies (x + 4)(x - 6) = 0$

$\implies (x + 4) = 0 \: \: or \: \: (x - 6) = 0$

$\implies x = - 4 \: \: or \: \: x = 6$

Then,

$\implies \: y = - 16 \: \: or \: \: y = 4$

So, the coordinates of point of intersection are

$(-4,-16)\:\:\:\&\:\:\:(6,4)$

Required distance

$AB = \sqrt{ {(6 + 4)}^{2} + (4 + 16) ^{2} }$

$\implies \: AB = \sqrt{ {(10)}^{2} + (20) ^{2} }$

$\implies \: AB = \sqrt{ 100 + 400 }$

$\implies \: AB = \sqrt{ 500 }$

$\implies \: AB = 10\sqrt{ 5}$