The equations of the tangents to the hyperbola $3 x^{2} - 4 y^{2} = 12$ which are parallel to the line 2x + y + 7 = 0 are

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Solution

The correct option is A
Equation of tangents are $y = m x \pm \sqrt{a^{2} m^{2} - b^{2}}$
Here m = slope of $2 x + y + 7 = 0 \Rightarrow m = - 2; a^{2} = 4; b^{2} = 3$ .
$∴ 2 x + y \pm \sqrt{13} = 0$

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Similar questions

3 x^{2} + 4 y^{2} = 12

2 x - y + 5 = 0

x = 3 t^{2}, y = 2 t^{3}

3 x^{2} + 4 y^{2} = 12

2 x - y + 5 = 0

3 x^{2} + 4 y^{2} = 12,

y + 2 x = 4,

2 x^{2} - 3 y^{2} = 6

y = 3 x + 4

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