k = 1/10
Learn how to use the discriminant of a quadratic equation when solving these types of questions.
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
Tangent:In geometry, the tangent line (or simply the tangent) is a curve at a given point and is the straight line that "just touches" the curve at that point. As it passes through the point where the tangent line and the curve meet the tangent line is "going in the same direction" as the curve, and in this sense it is the best straight-line approximation to the curve at that point.Chord:A chord of a curve is a geometric line segment whose endpoints both lie on the outside of the circle.
The tangent to a curve at a point refers to the straight line that best approaches the curve. In other words, in a small interval the straight line it will go in the same direction as the curve.
In geometry a straight line that touches a curve is called a tangent.
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
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.
Yes a tangent is a straight line thattouches a curve at only one point But there is a tangent ratio used in trigonometry
The answer depends on the context. In the context of a curve, a tangent is a straight line that touches the line without intersecting it. The antonym does not have a specific name because it could be a straight line that does not meet the curve at all, or it could be one that crosses the curve. Note that even a tangent can cross the curve at some other [distant] point(s).
It is a straight line that touches the curve such that the line is perpendicular to the radius of the curve at the point of contact.
You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4
equation 1: y = x-4 => y2 = x2-8x+16 when both sides are squared equation 2: x2+y2 = 8 Substitute equation 1 into equation 2: x2+x2-8x+16 = 8 => 2x2-8x+8 = 0 If the discriminant of the above quadratic equation is zero then this is proof that the line is tangent to the curve: The discriminant: b2-4ac = (-8)2-4*2*8 = 0 Therefore the discriminant is equal to zero thus proving that the line is tangent to the curve.
A tangent at that point where a straight line just touches a curve and a secant line when the straight line bisects the curve.
If the line y = x+c is a tangent to the curve y = 3x-x-5x^2 then the point of contact is made at (-0.2, 3) solved by using the discriminant and quadratic equation formulae
Usually a straight line that touches a curve at one point. At the point of contact, the tangent is perpendicular to the radius of curvature.
A straight line which is approached , but never reached by an infinite branch of a curve , and which can be regarded as a line tangent to the curve at infinity.
Combine the two equations together to give a quadratic equation in the form of:- 4x2 - 5x + 25/16 = 0 The solution to this is x = 5/8 or x = 5/8 meaning that it has equal roots therefore the line is tangent to the curve. The discriminant of b2 - 4ac = 0 also proves that the quadratic equation has two equal roots which makes the line tangent to the curve. Further proof can be found by plotting the straight line and curve graphically.
A tangent. And it does not need to be a circle - it can be any curve.
y=0. note. this is a very strange "curve". If y=0 then any value of x satisfies the equation, leading to a curve straight along the y axis. For any non-zero value of y the curve simplifies to y = -x. The curve is not differentiable at the origin.
The general form of a line tangent to a circle is:y=mx+a(1+m2 )1/2. where "a" is the radius of the circle. Here circle is x2 + y2=4, so radius=a=2. nowc2=a2(1+m2)=8 (given)or 8=4(1+m2)2=1+m2 orm2=1 orm=1. so equation becomesy=mx+c ory=x+cImproved Answer:Equation 1: y = x+square root of 8 => y2 = x2+square root of 32x+8 when both sides are squared.Equation 2: y2 = 4-x2By definition:x2+the square root of 32x+8 = 4-x2 => 2x2+the square root of 32x+4 = 0If the discriminant b2-4ac of the above quadratic equation is equal to zero then this is proof that the straight line is tangent to the curve:b2-4ac = the square root of 322-4*2*4 = 0Therefore the straight line is a tangent to the curve because the discriminant of the quadratic equation equals zero.
A tangent line is a line that is parallel to and intersecting a point on a curve, where the limit of the distance between two points on that curve (one of them being the point in question) and those two points also being on that line approaches zero. Since a straight line has only one slope, at all points on the line, then there can only be one tangent line to a straight line, and the tangent line is the same line as the straight line.
Considering an asymptote as a tangent to the curve "at infinity", the asymptote is the straight line itself.
Because a vector contains information about the direction. A direction, at any particular position is the tangent to the curve and this, by definition, must be straight.
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are: 1 Plot the given values on a graph and the line will touch the curve at one point 2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0 3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
You find the tangent to the curve at the point of interest and then find the slope of the tangent.