Find the equation of the tangent to the unit circle when x=sqrt(3)/2 (in the first quadrant)

Unit circle: x2 + y2 = 1 when x = sqrt(3)/2:  y2 = 1 - (sqrt(3)/2)2  y2 = 1 - 3/4  y2 = 1/4  y = 1/2 or -1/2 (first quadrant, so y is positive, i.e. y = 1/2) find gradient at (sqrt(3)/2, 1/2):  x2 + y2 = 1  2x + 2y dy/dx = 0  dy/dx = -2x/2y  dy/dx = -x/y Substitute x= sqrt(3)/2, y = 1/2  dy/dx = -sqrt(3) Find equation of line:  y - y1 = m(x - x1)  y - 1/2 = -sqrt(3)(x-sqrt(3)/2)  y = -sqrt(3)x + 3/2 + 1/2  y = -sqrt(3)x + 2

Answered by Kiran J. Maths tutor

3423 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The velocity of a car at time, ts^-1, during the first 20 s of its journey, is given by v = kt + 0.03t^2, where k is a constant. When t = 20 the acceleration of the car is 1.3ms^-2, what is the value of k?


How do I show two vectors are perpendicular?


A curve C has equation y = x^2 − 2x − 24x^(1/2) x > 0 find dy/dx


How do you know if a stationary point on a curve is a maximum or minimum without plotting the graph?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences