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Grade 12Differential Calculus

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

Profile image of taniska
11 Years agoGrade 12
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1 Answer

Profile image of Jitender Singh
11 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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