# Find the equation of the tangent at the point (x1.y1) to the curve x^m/a^m+y^m/b^m=1

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justaguide | Certified Educator

The curve we have is : x^m/a^m+y^m/b^m=1

If we differentiate the two sides we get:

(1/a^m)(m*x^(m-1)) + (1/b^m)(m*y^(m-1))dy/dx = 0

=> dy/dx = [-(1/a^m)(m*x^(m-1))]/[(1/b^m)(m*y^(m-1))]

The slope of the tangent to the the curve at the point (x1, y1) is the value of the first derivative at (x1 , y1) which is:

[-(1/a^m)(m*x1^(m-1))]/[(1/b^m)(m*y1^(m-1))]

The equation of the tangent passing through (x1, y1) is that of a line with a slope we have derived above passing through the point (x1, y1)

This is (y - y1)/(x - x1) = [-(1/a^m)(m*x1^(m-1))]/[(1/b^m)(m*y1^(m-1))]

**The required equation of the tangent is (y - y1)/(x - x1) = [-(1/a^m)(m*x1^(m-1))]/[(1/b^m)(m*y1^(m-1))]**