Find the equation in polar form of the line tangent to the unit circle

Question

at the point (-√3/2,-1/2)? 

 

 

Answer 1

The standard form of a circle equation is , where, center of the circle is (h ,k ) and radius of the circle is r .

A unit circle is a circle with radius of 1 and centered at the origin (0, 0) in the cartesian coordinate system.

So, the equation of unit circle is x ² + y ² = 1.

 

The unit circle is x ² + y ² = 1 and the point is (- √3/2, - 1/2).

Differentiate the unit circle with respect to x.

2x + 2yy ' = 0

x + yy ' = 0

y ' = - x /y.

When, x = - √3/2 and y = - 1/2, y ' = - (- √3/2 )/(- 1/2) = - √3

y ' = - √3.

This is the slope (m ) of the tangent line to the unit circle at (- √3/2, - 1/2).

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = - √3x  + b.

 

Find the y - intercept by substituting the the point in the tangent line equation say (x , y) = (- √3/2, - 1/2).

- 1/2 = (- √3)(- √3/2) + b

b = - 1/2 - 3/2

b = - 4/2

⇒ b = - 2.

The tangent line equation  is y = - √3x - 2.

 

The trigonometric functions cosine and sine may be defined for the linear equation.

If (x , y) is a point of a linear equation, and it makes the angle t from the positive x - axis, (where counterclockwise turning is positive), then,

x = r cos t and y = r sin t.

Slope (m ) = - √3

tan t = - √3

⇒ t = 2π/3.

Radius (r ) = 1.

x = 1 cos (2π/3) and y = 1 sin (2π/3).

So, the polar form of the tangent line equation is sin (2π/3) + √3(cos (2π/3)) = - 2.

Answer 2

The standard form of a circle equation is , where, center of the circle is (h ,k ) and radius of the circle is r .

A unit circle is a circle with radius of 1 and centered at the origin (0, 0) in the cartesian coordinate system.

So, the equation of unit circle is x ² + y ² = 1.

The unit circle is x ² + y ² = 1 and the point is (- √3/2, - 1/2).

Differentiate the unit circle with respect to x.

2x + 2yy ' = 0

x + yy ' = 0

y ' = - x /y.

When, x = - √3/2 and y = - 1/2, y ' = - (- √3/2 )/(- 1/2) = - √3

y ' = - √3.

This is the slope (m ) of the tangent line to the unit circle at (- √3/2, - 1/2).

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = - √3x  + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x , y) = (- √3/2, - 1/2).

- 1/2 = (- √3)(- √3/2) + b

b = - 1/2 - 3/2

b = - 4/2

⇒ b = - 2.

The tangent line equation  is y = - √3x - 2.

√3x + y = - 2

put, x = r cos θ and y = r sin θ.

√3(r cos θ ) + (r sin θ ) = - 2

r(√3 cos θ + sin θ ) = - 2

r = - 2/(√3 cos θ + sin θ ).

So, the polar form of the tangent line equation is r = - 2/(√3 cos θ + sin θ).