This is not possible, since the point (4,6) lies inside the circle : X2 + Y2 = 16 Tangents to a circle or ellipse never pass through the circle
The line of bisection of an ellipse is called the tangent.
If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.
(2, -2)
There is no sensible simplification. One possible answer is (tan + 1)*(tan - 1)
A line can have at most one root and so coincident roots for a line are not possible.
The line of bisection of an ellipse is called the tangent.
pineapple
In trig, the secant squared divided by the tangent equals the hypotenuse squared divided by the product of the opposite and adjacent sides of the triangle.Details: secant = hypotenuse/adjacent (H/A) and tangent = opposite/adjacent (A/O);Then secant2/tangent = (H2/A2)/(O/A) = H2/A2 x A/O = H2/AO.
If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.
The link below gives the equations for the line that tangent. Once you have that, finding the normal should be straightforward.http://mathworld.wolfram.com/CircleTangentLine.html
the derivative of tangent dy/dx [ tan(u) ]= [sec^(2)u]u' this means that the derivative of tangent of u is secant squared u times the derivative of u.
(2, -2)
It is: pi*0.875 squared*3 = 7.216 cubic m rounded to 3 d.p.
Yes, it does.
there could be many... sin2(a/2) / cos2(a/2)
There is no sensible simplification. One possible answer is (tan + 1)*(tan - 1)
k = 0.1