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What is the value of k when the line of y equals 2x plus k meets the curve of y equals x squared -2x -7 at one point showing work?
Wiki User
∙ 8y ago
If: y = x^2-2x=7 and y = 2x+y
Then: x^2-2x-7 = 2x+k => x^2-4x-(7-k) = 0
Using the discriminant of b^2-4(a*b) = 0
16-4(1(-7-k)) = 0 => 4k = -44 => k = -11
For the line to meet the curve at one point the discriminant must equal 0
Check: 16-4(1(-7--11)) = 0
Therefore the value of k = -11
Wiki User
∙ 8y ago
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What is the point of cotact when the line y equals 2x plus 5 over 4 meets the curve y squared equals 10xa?
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)
What is the length of the line y equals 8 -x that meets the curve of y equals x squared plus 4x plus 2 at two distinctive points?
7*sqrt(2) = 9.899 to 3 dp
What are the equal roots when y equals x -4 meets the curve of x squared plus y squared equals 8 showing work?
If: y = x-4 and x^2+y^2 = 8 then x^2+(x-4)^2 = 8 Multiply out the brackets and subtract 8 from both sides:- So: x^2+x^2-8x+16-8 = 0 => 2x^2-8x+8 = 0 Solving the above quadratic equation: x = 2 or x = 2 which are both equal roots
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?
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What is the length of the line y equals 5x plus 10 that meets the curve of y equals x squared plus 4 at two distinctive points?
The line of y = 5x+10 intersects with the curve of y = x2+4 at (6, 40) and (-1, 5) Length of the line is the square root of: (6--1)2+(40-5)2 = 7 times sq rt of 26 or 35.693 to three decimal places
What is the point of cotact when the line y equals 2x plus 5 over 4 meets the curve y squared equals 10xa?
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)
What is the length of the line y equals 8 -x that meets the curve of y equals x squared plus 4x plus 2 at two distinctive points?
7*sqrt(2) = 9.899 to 3 dp
What are the equal roots when y equals x -4 meets the curve of x squared plus y squared equals 8 showing work?
If: y = x-4 and x^2+y^2 = 8 then x^2+(x-4)^2 = 8 Multiply out the brackets and subtract 8 from both sides:- So: x^2+x^2-8x+16-8 = 0 => 2x^2-8x+8 = 0 Solving the above quadratic equation: x = 2 or x = 2 which are both equal roots
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 is the length of the line y equals 5x plus 10 that meets the curve of y equals x squared plus 4 at two distinctive points?
The line of y = 5x+10 intersects with the curve of y = x2+4 at (6, 40) and (-1, 5) Length of the line is the square root of: (6--1)2+(40-5)2 = 7 times sq rt of 26 or 35.693 to three decimal places
What are the coordinates of the straight line of y equals 5x plus 10 that meets the parabola of y equals x squared plus 4 at points A and B?
(6, 40) and (-1, 5)
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)
What is the perpendicular distance from 2 5 that meets the line of 7x plus 13 -y equals 0 at right angles showing work?
Equation: 7x+13-y = 0 => y = 7x+13Slope: 7Perpendicular slope: -1/7Perpendicular equation: 7y = -x+37Both equations intersect at: (-1.08, 5.44)Distance: square root of [(-1.08-2)squared+(5.44-5)squared)] = 3.111 to 3 d.p.
Where are the points of contact when the line x equals 2 -2y crosses the curve x squared plus 4y squared equals 4?
The two equations are:x = 2 - 2yx² + 4y² = 4The points where the first meets the second are the values of x and y that simultaneously satisfy both equations:Substitute using (1) for x in (2) and solve the formed quadratic:x² + 4y² = 4→ (2 - 2y)² + 4y² = 4→ 4 - 8y + 4y² + 4y² -4 = 0→ 8y² - 8y = 0→ 8y(y - 1) = 0→ y = 0 or y = 1Substituting these values back into (1) gives:y = 0 → x = 2 - 2× 0 = 2 - 0 = 2y = 1 → x = 2 - 2× 1 = 2 - 2 = 0→ The line meets the curve at the points (2, 0) and (0, 1)
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
What are the points of contact when the line x -y equals 2 meets the curve x squared -4y squared equals 5?
If: x-y = 2 then x^2 = (2+y)^2 If: x^2 -4y^2 = 5 then x^2 = 4y^2 +5 So: 4y^2 +5 = (2+y)^2 Expanding brackets and transposing terms: 3y^2 +1 -4y = 0 Factorizing: (3y-1)(y-1) = 0 => y = 1/3 or y = 1 Therefore by substitution points of contact are at: (7/3, 1/3) and (3, 1)
What is the length of the line y equals 17-3x that meets the curve of y equals x squared plus 2x-7 at two distinctive points showing work?
Clearly some sort of homework problem, so I'll give you hints.The lines:y = 17 - 3xy = x2 + 2x - 7meet where:17 - 3x = x2 + 2x - 7⇒ x2 + 5x - 24 = 0.Solving for x (the above will factorise giving 2 solutions for x) you can then find the corresponding y's (by putting the values of x into either of the two equations).With the (x, y) pairs you can then calculate the length of the line using:length = √((x1 - xo)2 + (y1 - yo)2)where the end points of the line (found above) are (xo, yo) and (x1, y1).
