Find the tangent line equations for every point on the unit circle
Answer 1
Emily Hall
To find the tangent line equations for every point on the unit circle, we start with the unit circle equation:
$ x^2 + y^2 = 1 $
Differentiate implicitly with respect to $x$ to find the slope:
$ 2x + 2y \x0crac{dy}{dx} = 0 $
Solve for $ \x0crac{dy}{dx} $:
$ \x0crac{dy}{dx} = -\x0crac{x}{y} $
At a point $ (a, b) $ on the unit circle, the slope of the tangent is:
$ m = -\x0crac{a}{b} $
The tangent line equation at $ (a, b) $ is:
$ y – b = -\x0crac{a}{b}(x – a) $
Multiply through by $ b $ to get:
$ b(y – b) = -a(x – a) $
Simplify to obtain the final equation of the tangent line:
$ ax + by = 1 $
Answer 2
William King
To find the tangent line equations for every point on the unit circle, we start with the unit circle equation:
$ x^2 + y^2 = 1 $
Differentiate implicitly with respect to $x$ to find the slope:
$ 2x + 2y x0crac{dy}{dx} = 0 $
Solve for $ x0crac{dy}{dx} $:
$ x0crac{dy}{dx} = -x0crac{x}{y} $
At a point $ (a, b) $ on the unit circle, the slope of the tangent is:
$ m = -x0crac{a}{b} $
The tangent line equation at $ (a, b) $ is:
$ y – b = -x0crac{a}{b}(x – a) $
Answer 3
Henry Green
To find the tangent line equations for every point on the unit circle, we start with the unit circle equation:
$ x^2 + y^2 = 1 $
Differentiate implicitly with respect to $x$ to find the slope:
$ 2x + 2y x0crac{dy}{dx} = 0 $
Solve for $ x0crac{dy}{dx} $:
$ x0crac{dy}{dx} = -x0crac{x}{y} $
At a point $ (a, b) $ on the unit circle, the slope of the tangent is:
$ m = -x0crac{a}{b} $
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