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Combine the equations together and using the quadratic equation formula it works out that the point of contact is at (5/8, 5/2)

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The coordinates of the point are (0.625, 2.5).

Q: What is the point of contact when the line y equals 2x plus 1.25 touches the curve y squared equals 10x?

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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)

If you mean a line of y = 2x+5/4 and a curve of y^2 = 10x Then it works out that the line touches the curve at: (5/8, 5/2)

If: y = x^2 -10x +13 and y = x^2 -4x +7 Then: x^2 -10x +13 = x^2 -4x +7 Transposing terms: -6x +6 = 0 => -6x = -6 => x = 1 Substituting the value of x into the original equations point of contact is at: (1, 4)

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.

k = 0.1

Related questions

It is (-0.3, 0.1)

If y = 2x +10 and y^2 = 10x then by forming a single quadratic equation and solving it the point of contact is made at (5/8, 5/2)

-2

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)

If you mean a line of y = 2x+5/4 and a curve of y^2 = 10x Then it works out that the line touches the curve at: (5/8, 5/2)

-2

If: y = x^2 -10x +13 and y = x^2 -4x +7 Then: x^2 -10x +13 = x^2 -4x +7 Transposing terms: -6x +6 = 0 => -6x = -6 => x = 1 Substituting the value of x into the original equations point of contact is at: (1, 4)

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.

(2, -2)

k = 0.1

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.

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)