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
KJ
Related Maths A Level answers
All answers ▸The curve y = 4x^2 + a/ x + 5 has a stationary point. Find the value of the positive constant a given that y-ordinate of the stationary point is 32.
Solve the equation 2cos2(x) + 3sin(x) = 3, where 0<x<=π
What is the smallest possible value of the integral ∫(x-a)^2 dx between 0 and 1 as a varies?
3. The point P lies on the curve with equation y=ln(x/3) The x-coordinate of P is 3. Find an equation of the normal to the curve at the point P in the form y = ax + b, where a and b are constants.