0
How do you prove that the line y equals 2x plus 1.25 is tangent to the curve y squared equals 10x?
Wiki User
∙ 13y ago
If the discriminant b2-4ac of the quadratic equation equals zero then it will have two equal roots meaning that the line is tagent to the curve.
So by implication: (2x+1.25)(2x+1.25) = 10x
4x2-5x+25/16 = 0
Hence use the discriminant of b2-4ac :-
(-5)2-4*4*25/16 = 0
Therefore the discriminant equals 0 so the line will be tangent to the curve.
In fact working out the equation gives x having two equal roots of 5/8
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
How do you prove that the line y equals x-4 is tangent to the curve of x squared plus y squared equals 8?
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.
How do you prove in 3 different ways that the line of y equals x -4 is tangent to the curve of x squared plus y squared equals 8?
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
How can you prove that the straight line of y equals x plus k is a tangent to the curve of x squared plus y squared equals 4 and only when k squared equals 8?
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
How can you prove in at least two ways that the straight line of y equals 2x plus 5 over 4 is a tangent to the curve of y square equals 10x?
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.
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
How do you prove that the line y equals 2x plus 1.25 is a tangent to the curve y squared equals 10x?
If: y = 2x+5/4 and y^2 = 10x Then: y^2 = (2x+5/4)^2 So: 4x^2 +5x +25/16 = 10x Transposing terms: 4x^2 -5x +25/16 = 0 A line is tangent to a curve when the discriminant: b^2 -4ac = 0 Thus: 5^2 -4*4*25/16 = 0 Therefore: y = 2x+1.25 is a tangent to the curve y^2 = 10x
How can you prove that 2 equations ...x squared plus y squared equals 25... and a linear equation ...y equals -.75x plus 6.. are tangent to each other algebraically and without a calculator?
The solutions (x,y) to the first equation form the circle centered at 0, with radius 5 (the square root of 25). Solving for y, you can see that it consists of two functions of x: y = sqrt(25-x^2), y = -sqrt(25-x^2). Now you must show that y = -0.75x + 6 is tangent to one of them. That is, show that for one of the curves, at some point x', the slope is -0.75, AND the line intersects the curve at x' (that is, the y-values are equal for the curve and line, at point x'). Compute the derivative of each curve, and find for what x' it is equal to -0.75. (For each derivative, there is an easy integer solution which will work. No calculator needed.) Then check if the line intersects the curve at the point x' you find (for each curve). (It will work for one of the curves.) So you just showed that at point x', one of the curves has slope -0.75, and the line (also with slope -0.75) intersects the curve at x'. Therefore, the line is tangent to the curve at point x'. Hope that helps.
How do you prove that the straight line y equals x plus c is a tangent to the curve x squared plus y squared equals 4 when c squared equals 8?
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.
How do you prove that the line y equals 2x plus 5 over 4 touches the curve y squared equals 10x?
Y2=10X and Y=2x+5 will never touch each other. If there is a touch point then there will be a common value (Touch point) of x and y which will satisfy both equations. But there is no common point so it is not possibleImproved Answer:equation 1: y = 2x++5/4 => y2 = 4x2+5x+1.5625 when both sides are squaredequation 2: y2 = 10xBy definition: 4x2+5x+1.5625 = 10x => 4x2-5x+1.5625 = 0If the discriminant b2-4ac is equal to zero then the line is tangent to the curve:b2-4ac = (-5)2-4*4*1.5625 = 0Therefore the discriminant is zero thus proving that the line is tangent to the curve.
Prove identity sin4x equals 4sinxcosx times 1 minus 2sin squared x?
sin4x=(4sinxcosx)(1-2sin^2x)
Who discovered E equals MC squared?
It is believed that Einstein was the one who made up the equation yet there is much speculation surrounding it validity and many have tried to prove him wrong.
How can I prove that 2x squared -11x-21 is an acute angle?
I am not sure I understand what you are asking. The formula you give is a parabola and the gradient of the curve is changing all the time. So please can you give more details about where this acute angle might be.
