Find the equation of tangents to the curve ${3 x}^{2} - y^{2} = 8$ which passes through the point $(\frac{4}{3}, 0)$

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Solution

$3 x^{2} - y^{2} = 8$
Slope of tangent at $(x, y)$ (point of contact) will be $\frac{3 x}{y}$ ….. $(1)$ (by differentiating the curve equation)
Tangent also passes through $(\frac{4}{3}, 0)$ so slope will be $\frac{y - 0}{x - \frac{4}{3}} \dots . (2)$
Equating $(1)$ (2)$
We get $x = 2, y = 2, - 2$
So equation of tangents will be $y = 3 x - 4, y = - 3 x + 4$

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Find the eqns of the tangent to curve 3x²-y²=8 which passes through the point (4/3,0).

3 x^{2} - y^{2} = 8

(\frac{4}{3}, 0)

y^{2} - 3 x^{2} - 4 y + 8 = 0

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(x, y)

(3, 0)

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