Hyperbola?! Find the standard form of the equation of the hyperbola whose graph is given below?
The center of the hyperbola is at the center of the dotted rectangle, which is at the origin. Your twin parabolas (which actually constitute the hyperbola) open to the left and right, so the major axis lies along the x-axis. The general form of this hyperbola is (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1 where (h,k) = (0,0) is the center, a=4, and b=2. (a and b are the distances from the center to the sides of the rectangle.) So your hyperbola is x^2 / 4^2 - y^2 / 2^2 = x^2 / 16 - y^2 / 4 = 1 (Answer) Notice that when y=0, x = ±4, so the vertices are at (-4, 0) and (4, 0). Extra note: If the "parabolas" opened up and down rather than left and right, the major axis would've been vertical, and the equation would be y^2 / 4 - x^2 / 16 = 1 with vertices at (0, 2) and (0, -2).
True or false 1- _____ A hyperbola will never intersect its transverse axis.?
1- _____ A hyperbola will never intersect its transverse axis. 2- _____ A circle is a certain type of ellipse. 3- _____ The foci, vertices, and center of an ellipse lie on a line called the axis of symmetry. 4- _____ The vertex of a parabola is a point on the parabola that also is on its axis of symmetry 5_____ The graph of a cubic function is a parabola.
Graph the hyperbola x^2/25 - y^2/49=1. What are the equations of the asymptotes?
Graph the hyperbola x^2/25 - y^2/49=1. What are the equations of the asymptotes? a) y= +/- (25/49)x b) y= +/- (5/7)x c) y= +/- (7/5)x d) y= +/- (49/25)x How do you solves this? Thanks in advance.
What is the equation of the hyperbola through the point V?
I assume that was your question I just finished answering before (note my answer is now edited there). I don't believe "the hyperbola" can be found. There's heaps of them (infinitely many). I missed that x intercept initially too when i first answered this question here's my answer from before: ok, so we have a straight line and parabola with a point in common on the x axis (x1, 0) Vertex is always half way between roots of parabola so since one root is at x = 1 and the vertex is at x = 3 the other root is at x = 5 point in common on the x axis (5, 0) straight line is: y - 0 = (4 - 0)/(-3 - 5)*(x - 5) y = -1/2*(x - 5) y = -x/2 + 5/2 parabola is: a(y - 2) = (x - 3)² sub point (1, 0) a(0 - 2) = (1 - 3)² -2a = 4 a = -2 -2(y - 2) = (x - 3)² -2y + 4 = (x - 3)² y = 2 - (x - 3)²/2 y = 2 - (x² - 6x + 9)/2 y = 2 - x²/2 + 3x - 9/2 y = - x²/2 + 3x - 5/2 distance between ST and T is d = (- x²/2 + 3x - 5/2) - (-x/2 + 5/2) = -x²/2 + 3x - 5/2 + x/2 - 5/2 = -x²/2 + 7x/2 - 5 d' = -x + 7/2 = 0 x = 7/2 = 3.5 d = -(3.5)²/2 + 7(3.5)/2 - 5 = 9/8 units (1.125) P is intercept of parabola and line - x²/2 + 3x - 5/2 = -x/2 + 5/2 x² - 6x + 5 = x - 5 x² - 7x + 10 = 0 (x - 5)(x - 2) = 0 its x = 2 since x = 5 is the other one already found at x = 2, y = -2/2 + 5/2 = 3/2 (1.5) distance from P to origin is √((3/2)² + 2²) = √(9/4 + 4) = √(25/4) = 5/2 units circle is: (x - 2)² + (y - 3/2)² = 25/4 for question 2.5 we need more info. Too hard to graph online
How do I find the intersection point of a line y=mx+c and a line parallel to the y axis? They both will definitely intersect (slope of line parallel to y axis is ∞).
Let there be an arbitrary line [math]y = mx + c[/math]Let it intersect a line parallel to the y-axis, whose equation is [math]x = a[/math]If we rearrange the first equation, we get:[math]x = \dfrac{y-c}{m}[/math]Solving these two equations,[math]\dfrac{y-c}{m} = a[/math][math]y = ma + c[/math]The point at which the lines will intersect is given as [math](a, ma+c)[/math]
Which of the following equations could be that of the hyperbola graphed below?
vertex info: (-3,3) means that no matter what conic section two things will appear: (x - -3) and (y - 3), so actually look for (x + 3) and (y - 3) This eliminates c, and leaves a) and b). They are BOTH hyperbolas, but one is vertical (reflected about the x-axis) and the other is horizontal (reflected about the y-axis). I don't know which is which so I have to figure it out and hopefully if you forget this will help you remember: Can you make x = 0?, you WON'T be able to if it's horizontal, think about it, if x = 0, then your point will lie on the y-axis (please please think about, I know it sounds weird). Does the y-axis intersect the hyperbola? Think about this, only a vertical hyperbola will intersect the y-axis (and therefore x can equal 0). Likewise can y = 0, then the point lies on the x-axis (only a horizontal hyperbola will have these points) OK, so if you make the x-part 0, i.e. (x+3) = 0. DON'T solve the eq. it's NOT necessary, only to know that that part will be zero and therefore not part of the eq. a.) gives you -y^2 = N (where A is a positive real number) therefore you have y^2 = -N AND therefore y does not exist, that is to say that when you set the x part to zero you CANNOT solve the equation for the y part and therefore you cannot set the x part to zero. On the other hand, if you try y-3 = 0 (set the y-part to zero) then you have the same type of equation (except your are now solving for x): x^2 = N, now you can take the square root and in fact get two solutions: x = +/- sqrt(N) (the left and right side of hyperbola) b.) same analysis, and since x and y are switched (in order) you know that you will find the opposite true...that is that you CAN set the x-part to zero (and get 2 solutions for y) and you will NOT be able to set the y-part to zero. therefore a) is a horizontal hyperbola and b) is a vertical hyperbola
"Line intersect parabola at single point." Does that mean the line is tangent?
NoIn fact, a tangent to a curve (parabola in this case) is defined as the line that intersects the curve at 2 infinitesimally close points.If I'm not wrong the only line that intersects the parabola at only 1 point is the axis of the parabola.Also if you define a tangent in the usual way (i.e. A line that touches a curve at a point) then it just touches the curve and not intersects it. So yeah, the axis of the parabola is the only line that intersects it at a single pointEDIT : I just saw one of the collapsed answers and realised that any diameter to the parabola intersects it at only 1 point.