The eccentricity of the conic section represented by the equation ${(10 x - 5)}^{2} + {(10 y - 5)}^{2} = {(3 x + 4 y - 1)}^{2}$ is

\frac{1}{2}

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\frac{1}{\sqrt{2}}

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\sqrt{2}

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\frac{1}{10}

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Solution

The correct option is A $\frac{1}{2}$

Given: $(10 x - 5)^{2} + (10 y - 5)^{2} = (3 x + 4 y - 1)^{2}$

$= 100 x^{2} + 25 - 100 x + 100 y^{2} + 25 - 100 y = (3 x + 4 y)^{2} + 1 - 2 (3 x + 4 y)$

$= 100 x^{2} + 100 y^{2} - 100 x - 100 y + 50 = 9 x^{2} + 16 y^{2} + 24 x y + 1 - 6 x - 8 y$

$= 91 x^{2} + 84 y^{2} - 94 x - 92 y - 24 x y + 49 = 0$

$= 91 x^{2} + 84 y^{2} - 24 x y - 94 x - 92 y + 49 = 0$

Compare it with $a x^{2} + B x y + C y^{2} + D x + E y + F = 0$

$A = 91 B = 24 C = 84 D = 94 E = - 92 F = 49$

$e = \sqrt{\frac{2 (A - C)^{2} + B^{2}}{(A + C) + \sqrt{(A - C)^{2} + B^{2}}}} = \sqrt{\frac{2 \times \sqrt{7^{2} + 24^{2}}}{175 + \sqrt{7^{2} + 24^{2}}}}$

$e = \sqrt{\frac{50}{200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

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(10 x - 5)^{2} + (10 y - 5)^{2} = (3 x + 4 y - 1)^{2}

(10 x - 5)^{2} + (10 y - 4)^{2} = λ^{2} (3 x + 4 y - 1)^{2}

3 x + 4 y - 5 = 0

x^{2} + y^{2} = (3 x + 4 y - 5)^{2}

3 (x + y - 5)^{2} + 2 (x - y + 7)^{2} = 6

(10 x - 5)^{2} + (10 y - 2)^{2} = 9 (3 x + 4 y - 7)^{2}

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