answersLogoWhite

0


Best Answer

If: x-2y = 1

Then: x^2 = 4y^2 +4y+1

If: 4y^2 -3x^2 = 1

Then: 4y^2 -3(4y^2 +4y+1) = 1

Removing brackets: 4y^2 -12y^2 -12y -3 = 1

Transposing terms: -8y^2 -12y -4 = 0

Dividing all terms by -4: 2y^2 +3y +1 = 0

Factorizing the above: (2y+1)(y+1) = 0 meaning y = -1/2 or -1

By substitution into original linear equation points of contact are at: (0, -1/2) and (-1, -1)

User Avatar

Wiki User

6y ago
This answer is:
User Avatar
More answers
User Avatar

Wiki User

6y ago

They are (-1, -1) and (0, -1/2).

This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: What are the points of contact between the line x -2y equals 1 and the curve 4y squared -3x squared equals 1?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What are the points of contact between the line 3x -2y equals 1 and the curve 3x squared -2y squared plus 5 equals 0?

Equations: 3x-2y = 1 and 3x^2 -2y^2 +5 = 0 By combining the equations into one single quadratic equation and solving it the points of contact are made at (3, 4) and (-1, -2)


What are the points of intersection of the line x -y equals 2 with x squared -4y squared equals 5?

The points of intersection are: (7/3, 1/3) and (3, 1)


What are the points of contact that the curve of y equals x squared -x -12 makes with the x and y axes on the Cartesian plane?

If: y = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)


What are the points of intersection of 3x -2y -1 equals 0 with 3x squared -2y squared equals -5 on the Cartesian plane?

Points of intersection work out as: (3, 4) and (-1, -2)


Where are the points of contact when the line 3x -y equals 5 passes through the curve 2x squared plus y squared equals 129?

It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)


What are the points of contact between the line 2y -x equals 0 and the curve x squared plus y squared equals 20?

If: 2y-x = 0 Then: 4y^2 = x^2 If: x^2 +y^2 = 20 Then: 4y^2 +y^2 = 20 So: 5y^2 = 20 Dividing both sides by 5: y^2 = 4 Square root of both sides: y = - 2 or + 2 By substituting points of contact are at: (4, 2) and (-4, -2)


What are the points of contact when the line 2x plus 5y equals 4 transverses the curve y squared equals x plus 4?

If 2x+5y = 4 and y^2 = x+4 then by combining the equations into a single quadratic equation and solving it the points of contact are made at (12, -4) and (-7/2, 3/2)


What are the possible points of contact when the line 2y -x equals 0 meets the curve x squared plus y squared equals 20?

If: 2y-x = 0 then 4y^2 = x^2 So : 4y^2 +y^2 = 20 or 5y^2 = 20 or y^2 = 4 Square rooting both sides: y = -2 or y = 2 Therefore possible points of contact are at: (4, 2) and (-4. -2)


Where are the points of contact when the line 2x plus 5 equals 4 crosses the curve y squared equals x plus 4 on the Cartesian plane?

The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))


Where are the points of intersection of the equations 4y squared -3x squared equals 1 and x -2y equals 1?

The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)


What are the full range of values for k when the line y equals x -3 meets the curve of x squared -3y squared equals k showing work?

0


What points of contact between bones?

joints