find the tangent to the ellipse £¨x^2/4)+(y^2/4£©=1 at the point
(1,rad(3)).
To find the tangent to the ellipse, we need to find the slope of the tangent line at the given point. The slope of a tangent line to a curve can be found using calculus by taking the derivative of the equation of the curve.
The equation of the ellipse is (x^2/4) + (y^2/4) = 1. To find the slope of the tangent line, we will differentiate both sides of the equation with respect to x.
Differentiating (x^2/4) + (y^2/4) = 1 with respect to x:
d/dx[(x^2/4) + (y^2/4)] = d/dx[1]
(1/4)(2x) + (1/4)(2y(dy/dx)) = 0
Simplifying the equation:
x/2 + y(dy/dx)/2 = 0
Now we need to find the value of dy/dx at the given point (1, √3). To do this, we will substitute x = 1 and y = √3 into the equation:
1/2 + (√3)(dy/dx)/2 = 0
Now we can solve for dy/dx:
(√3)(dy/dx)/2 = -1/2
dy/dx = (-1/2)(2/√3)
dy/dx = -1/√3
So the slope of the tangent line at the point (1, √3) is -1/√3.
Now, we have the slope of the tangent line and the point (1, √3). We can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values, we have:
y - √3 = (-1/√3)(x - 1)
Simplifying the equation further:
y - √3 = (-x + 1)/√3
Multiplying both sides of the equation by √3:
√3(y - √3) = -x + 1
√3y - 3 = -x + 1
Moving the terms around:
x + √3y = 4
So the equation of the tangent line to the ellipse at the point (1, √3) is x + √3y = 4.