Determine the coordinates of a point on the unit circle where the tangent line has a slope of $frac{3}{4}$

To determine the coordinates of a point on the unit circle where the tangent line has a slope of $\frac{3}{4}$, we start with the equation of the unit circle:

$x^2 + y^2 = 1$

The slope of the tangent line at a point $(x, y)$ on the circle can be found by differentiating implicitly:

$2x + 2y\frac{dy}{dx} = 0$

Solving for $\frac{dy}{dx}$, we get:

$\frac{dy}{dx} = -\frac{x}{y}$

We need the slope to equal $\frac{3}{4}$:

$-\frac{x}{y} = \frac{3}{4}$

This implies:

$y = -\frac{4x}{3}$

Substitute $y = -\frac{4x}{3}$ back into the unit circle equation:

$x^2 + \left(-\frac{4x}{3}\right)^2 = 1$

$x^2 + \frac{16x^2}{9} = 1$

$\frac{25x^2}{9} = 1$

$x^2 = \frac{9}{25}$

$x = \pm \frac{3}{5}$

Substitute $x$ back into $y = -\frac{4x}{3}$:

$y = \mp \frac{4 \cdot \frac{3}{5}}{3} = \mp \frac{4}{5}$

The coordinates are:

$\left(\frac{3}{5}, -\frac{4}{5}\right)$ and $\left(-\frac{3}{5}, \frac{4}{5}\right)$

Find points on the unit circle where the tangent line has a slope of $frac{3}{4}$.

The unit circle equation is:

$x^2 + y^2 = 1$

Implicit differentiation gives:

$2x + 2yfrac{dy}{dx} = 0 $

The slope $frac{dy}{dx} $:

$ frac{dy}{dx}= -frac{x}{y} $

Set slope $frac{dy}{dx} $ equal $frac{3}{4}$:

$ – frac{x}{y} = frac{3}{4} $

Thus,:

$ y= – frac{4x}{3} $

Substitute in unit circle equation:

$ x^2 + left( – frac{4x}{3}
ight)^2 = 1 $

$ x^2 + frac{16 x^2}{9}= 1 $

$ frac{25 x^2}{9} = 1 $

$ x^2 = frac{9}{25} $

$ x = pm frac{3}{5} $

Substitute back in for $ y $ gives:

$ y = mp frac{4}{5} $

Hence points are:

$ left( frac{3}{5}, – frac{4}{5}
ight) $ and $ left( – frac{3}{5}, frac{4}{5}
ight) $

Given unit circle:

$x^2 + y^2 = 1$

Differentiate implicitly:

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