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tangents and normals
find the equation of the tangents to the curve 3x^@ – y^2 = 8 which passes through the point (4/3 , 0)

taniska , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$3x^2 - y^2 = 8$
This is the equation of hyperbola.
Equation of tangent in slope (m) form:
$y = mx \pm \sqrt{\frac{8}{3}m^2 - 8}$
It is passing through (4/3, 0)
$0 = \frac{4}{3}m \pm \sqrt{\frac{8}{3}m^2 - 8}$
$\frac{16}{9}m^2 = \frac{8}{3}m^2 - 8$
$\frac{16}{9}m^2 = \frac{24}{9}m^2 - 8$
$\frac{8}{9}m^2 = 8$
$m = \pm 3$
$y = \pm 3x\pm 4$

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