hyperbola word problems with solutions and graph

\(\dfrac{{(x3)}^2}{9}\dfrac{{(y+2)}^2}{16}=1\). Round final values to four decimal places. See Example \(\PageIndex{4}\) and Example \(\PageIndex{5}\). A hyperbola is a set of all points P such that the difference between the distances from P to the foci, F1 and F2, are a constant K. Before learning how to graph a hyperbola from its equation, get familiar with the vocabulary words and diagrams below. by b squared. The length of the latus rectum of the hyperbola is 2b2/a. }\\ c^2x^2-2a^2cx+a^4&=a^2(x^2-2cx+c^2+y^2)\qquad \text{Expand the squares. Direct link to King Henclucky's post Is a parabola half an ell, Posted 7 years ago. And once again, just as review, is equal to r squared. vertices: \((\pm 12,0)\); co-vertices: \((0,\pm 9)\); foci: \((\pm 15,0)\); asymptotes: \(y=\pm \dfrac{3}{4}x\); Graphing hyperbolas centered at a point \((h,k)\) other than the origin is similar to graphing ellipses centered at a point other than the origin. this when we actually do limits, but I think Its equation is similar to that of an ellipse, but with a subtraction sign in the middle. Direct link to khan.student's post I'm not sure if I'm under, Posted 11 years ago. The rest of the derivation is algebraic. You have to distribute y = y\(_0\) (b / a)x + (b / a)x\(_0\) over a squared x squared is equal to b squared. See you soon. If each side of the rhombus has a length of 7.2, find the lengths of the diagonals. The equation of the hyperbola can be derived from the basic definition of a hyperbola: A hyperbola is the locus of a point whose difference of the distances from two fixed points is a constant value. This difference is taken from the distance from the farther focus and then the distance from the nearer focus. line and that line. bit more algebra. So this number becomes really So that's a negative number. b, this little constant term right here isn't going There was a problem previewing 06.42 Hyperbola Problems Worksheet Solutions.pdf. Solve for the coordinates of the foci using the equation \(c=\pm \sqrt{a^2+b^2}\). same two asymptotes, which I'll redraw here, that 7. Direct link to Alexander's post At 4:25 when multiplying , Posted 12 years ago. of the other conic sections. Anyway, you might be a little Thus, the vertices are at (3, 3) and ( -3, -3). So we're not dealing with Graph the hyperbola given by the equation \(\dfrac{x^2}{144}\dfrac{y^2}{81}=1\). get rid of this minus, and I want to get rid of Solutions: 19) 2212xy 1 91 20) 22 7 1 95 xy 21) 64.3ft When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. And then you could multiply Direct link to ryanedmonds18's post at about 7:20, won't the , Posted 11 years ago. one of these this is, let's just think about what happens Use the standard form \(\dfrac{{(xh)}^2}{a^2}\dfrac{{(yk)}^2}{b^2}=1\). Further, another standard equation of the hyperbola is \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\) and it has the transverse axis as the y-axis and its conjugate axis is the x-axis. If a hyperbola is translated \(h\) units horizontally and \(k\) units vertically, the center of the hyperbola will be \((h,k)\). When we slice a cone, the cross-sections can look like a circle, ellipse, parabola, or a hyperbola. This could give you positive b Looking at just one of the curves: any point P is closer to F than to G by some constant amount. 2a = 490 miles is the difference in distance from P to A and from P to B. If the signal travels 980 ft/microsecond, how far away is P from A and B? One, because I'll We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: \(\dfrac{x^2}{a^2}\dfrac{y^2}{b^2}=1\), where the branches of the hyperbola form the sides of the cooling tower. look something like this, where as we approach infinity we get Approximately. Fancy, huh? Solve applied problems involving hyperbolas. Challenging conic section problems (IIT JEE) Learn. or minus square root of b squared over a squared x \[\begin{align*} d_2-d_1&=2a\\ \sqrt{{(x-(-c))}^2+{(y-0)}^2}-\sqrt{{(x-c)}^2+{(y-0)}^2}&=2a\qquad \text{Distance Formula}\\ \sqrt{{(x+c)}^2+y^2}-\sqrt{{(x-c)}^2+y^2}&=2a\qquad \text{Simplify expressions. Auxilary Circle: A circle drawn with the endpoints of the transverse axis of the hyperbola as its diameter is called the auxiliary circle. by b squared, I guess. Answer: Asymptotes are y = 2 - ( 3/2)x + (3/2)5, and y = 2 + 3/2)x - (3/2)5. It just gets closer and closer Notice that the definition of a hyperbola is very similar to that of an ellipse. You write down problems, solutions and notes to go back. those formulas. square root, because it can be the plus or minus square root. Let's see if we can learn squared over a squared x squared plus b squared. Making educational experiences better for everyone. Which axis is the transverse axis will depend on the orientation of the hyperbola. cancel out and you could just solve for y. The eccentricity of a rectangular hyperbola. Solution: Using the hyperbola formula for the length of the major and minor axis Length of major axis = 2a, and length of minor axis = 2b Length of major axis = 2 6 = 12, and Length of minor axis = 2 4 = 8 always forget it. answered 12/13/12, Certified High School AP Calculus and Physics Teacher. Length of major axis = 2 6 = 12, and Length of minor axis = 2 4 = 8. If the equation is in the form \(\dfrac{x^2}{a^2}\dfrac{y^2}{b^2}=1\), then, the coordinates of the vertices are \((\pm a,0)\0, If the equation is in the form \(\dfrac{y^2}{a^2}\dfrac{x^2}{b^2}=1\), then. you get b squared over a squared x squared minus What is the standard form equation of the hyperbola that has vertices \((0,\pm 2)\) and foci \((0,\pm 2\sqrt{5})\)? So in the positive quadrant, center: \((3,4)\); vertices: \((3,14)\) and \((3,6)\); co-vertices: \((5,4)\); and \((11,4)\); foci: \((3,42\sqrt{41})\) and \((3,4+2\sqrt{41})\); asymptotes: \(y=\pm \dfrac{5}{4}(x3)4\). The standard equation of the hyperbola is \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) has the transverse axis as the x-axis and the conjugate axis is the y-axis. original formula right here, x could be equal to 0. All rights reserved. confused because I stayed abstract with the By definition of a hyperbola, \(d_2d_1\) is constant for any point \((x,y)\) on the hyperbola. My intuitive answer is the same as NMaxwellParker's. I will try to express it as simply as possible. Graphing hyperbolas (old example) (Opens a modal) Practice. you've already touched on it. Direct link to RoWoMi 's post Well what'll happen if th, Posted 8 years ago. Determine which of the standard forms applies to the given equation. immediately after taking the test. And then since it's opening Find the equation of a hyperbola with foci at (-2 , 0) and (2 , 0) and asymptotes given by the equation y = x and y = -x. An hyperbola looks sort of like two mirrored parabolas, with the two halves being called "branches". squared minus b squared. Therefore, \(a=30\) and \(a^2=900\). The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. But it takes a while to get posted. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. If the plane intersects one nappe at an angle to the axis (other than 90), then the conic section is an ellipse. Using the reasoning above, the equations of the asymptotes are \(y=\pm \dfrac{a}{b}(xh)+k\). So we're going to approach squared minus x squared over a squared is equal to 1. The length of the transverse axis, \(2a\),is bounded by the vertices. What does an hyperbola look like? Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. Graph hyperbolas not centered at the origin. Find the equation of the parabola whose vertex is at (0,2) and focus is the origin. In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. give you a sense of where we're going. Solution. . The distance from P to A is 5 miles PA = 5; from P to B is 495 miles PB = 495. So as x approaches infinity. So once again, this And what I like to do If the plane is perpendicular to the axis of revolution, the conic section is a circle. re-prove it to yourself. It doesn't matter, because So that tells us, essentially, The vertices are located at \((0,\pm a)\), and the foci are located at \((0,\pm c)\). complicated thing. So as x approaches positive or To solve for \(b^2\),we need to substitute for \(x\) and \(y\) in our equation using a known point. Choose an expert and meet online. Since c is positive, the hyperbola lies in the first and third quadrants. when you take a negative, this gets squared. 9x2 +126x+4y232y +469 = 0 9 x 2 + 126 x + 4 y 2 32 y + 469 = 0 Solution. equal to 0, but y could never be equal to 0. over a squared to both sides. I have a feeling I might And that is equal to-- now you Direct link to akshatno1's post At 4:19 how does it becom, Posted 9 years ago. WORD PROBLEMS INVOLVING PARABOLA AND HYPERBOLA Problem 1 : Solution : y y2 = 4.8 x The parabola is passing through the point (x, 2.5) satellite dish is More ways to get app Word Problems Involving Parabola and Hyperbola then you could solve for it. In the next couple of videos The vertices and foci are on the \(x\)-axis. Since the speed of the signal is given in feet/microsecond (ft/s), we need to use the unit conversion 1 mile = 5,280 feet. And what I want to do now is It follows that \(d_2d_1=2a\) for any point on the hyperbola. You have to do a little little bit lower than the asymptote, especially when Example Question #1 : Hyperbolas Using the information below, determine the equation of the hyperbola. The sum of the distances from the foci to the vertex is. You can set y equal to 0 and y 2 = 4ax here a = 1.2 y2 = 4 (1.2)x y2 = 4.8 x The parabola is passing through the point (x, 2.5) (2.5) 2 = 4.8 x x = 6.25/4.8 x = 1.3 m Hence the depth of the satellite dish is 1.3 m. Problem 2 : An engineer designs a satellite dish with a parabolic cross section. The difference 2,666.94 - 26.94 = 2,640s, is exactly the time P received the signal sooner from A than from B. We can observe the different parts of a hyperbola in the hyperbola graphs for standard equations given below. For any point on any of the branches, the absolute difference between the point from foci is constant and equals to 2a, where a is the distance of the branch from the center. You couldn't take the square The standard form of the equation of a hyperbola with center \((h,k)\) and transverse axis parallel to the \(x\)-axis is, \[\dfrac{{(xh)}^2}{a^2}\dfrac{{(yk)}^2}{b^2}=1\]. huge as you approach positive or negative infinity. We know that the difference of these distances is \(2a\) for the vertex \((a,0)\). Let me do it here-- The hyperbola has only two vertices, and the vertices of the hyperbola \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) is (a, 0), and (-a, 0) respectively. Interactive online graphing calculator - graph functions, conics, and inequalities free of charge Note that the vertices, co-vertices, and foci are related by the equation \(c^2=a^2+b^2\). Using the one of the hyperbola formulas (for finding asymptotes): Well what'll happen if the eccentricity of the hyperbolic curve is equal to infinity? as x becomes infinitely large. Thus, the transverse axis is on the \(y\)-axis, The coordinates of the vertices are \((0,\pm a)=(0,\pm \sqrt{64})=(0,\pm 8)\), The coordinates of the co-vertices are \((\pm b,0)=(\pm \sqrt{36}, 0)=(\pm 6,0)\), The coordinates of the foci are \((0,\pm c)\), where \(c=\pm \sqrt{a^2+b^2}\). An hyperbola is one of the conic sections. We're going to add x squared Direct link to superman's post 2y=-5x-30 The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. The distance of the focus is 'c' units, and the distance of the vertex is 'a' units, and hence the eccentricity is e = c/a. Breakdown tough concepts through simple visuals. The design efficiency of hyperbolic cooling towers is particularly interesting. Solve for \(b^2\) using the equation \(b^2=c^2a^2\). could never equal 0. Calculate the lengths of first two of these vertical cables from the vertex. circle and the ellipse. It was frustrating. over a squared plus 1. The crack of a whip occurs because the tip is exceeding the speed of sound. y = y\(_0\) - (b/a)x + (b/a)x\(_0\) and y = y\(_0\) - (b/a)x + (b/a)x\(_0\), y = 2 - (4/5)x + (4/5)5 and y = 2 + (4/5)x - (4/5)5. }\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\qquad \text{Rearrange terms. }\\ cx-a^2&=a\sqrt{{(x-c)}^2+y^2}\qquad \text{Divide by 4. Hyperbola Word Problem. And if the Y is positive, then the hyperbolas open up in the Y direction. of Important terms in the graph & formula of a hyperbola, of hyperbola with a vertical transverse axis. These are called conic sections, and they can be used to model the behavior of chemical reactions, electrical circuits, and planetary motion. asymptote will be b over a x. Solution : From the given information, the parabola is symmetric about x axis and open rightward. Solving for \(c\), \[\begin{align*} c&=\sqrt{a^2+b^2}\\ &=\sqrt{49+32}\\ &=\sqrt{81}\\ &=9 \end{align*}\]. This page titled 10.2: The Hyperbola is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. my work just disappeared. it if you just want to be able to do the test If it is, I don't really understand the intuition behind it. As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. divided by b, that's the slope of the asymptote and all of Because in this case y Also, what are the values for a, b, and c? x approaches infinity, we're always going to be a little That this number becomes huge. Now you said, Sal, you This number's just a constant. Minor Axis: The length of the minor axis of the hyperbola is 2b units. Sal introduces the standard equation for hyperbolas, and how it can be used in order to determine the direction of the hyperbola and its vertices. Find the required information and graph: . hyperbolas, ellipses, and circles with actual numbers. Parametric Coordinates: The points on the hyperbola can be represented with the parametric coordinates (x, y) = (asec, btan). And since you know you're Center of Hyperbola: The midpoint of the line joining the two foci is called the center of the hyperbola. Or our hyperbola's going They look a little bit similar, don't they? Round final values to four decimal places. This is because eccentricity measures who much a curve deviates from perfect circle. For problems 4 & 5 complete the square on the x x and y y portions of the equation and write the equation into the standard form of the equation of the ellipse. asymptotes look like. always use the a under the positive term and to b Legal. Find the eccentricity of x2 9 y2 16 = 1. A rectangular hyperbola for which hyperbola axes (or asymptotes) are perpendicular or with an eccentricity is 2. And in a lot of text books, or The transverse axis is along the graph of y = x. }\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{{(x-c)}^2+y^2}+x^2-2cx+c^2+y^2\qquad \text{Expand remaining square. Major Axis: The length of the major axis of the hyperbola is 2a units. The slopes of the diagonals are \(\pm \dfrac{b}{a}\),and each diagonal passes through the center \((h,k)\). If you multiply the left hand The coordinates of the vertices must satisfy the equation of the hyperbola and also their graph must be points on the transverse axis. As a helpful tool for graphing hyperbolas, it is common to draw a central rectangle as a guide. to get closer and closer to one of these lines without A hyperbola is the locus of a point whose difference of the distances from two fixed points is a constant value. Graph the hyperbola given by the equation \(9x^24y^236x40y388=0\). away, and you're just left with y squared is equal Substitute the values for \(h\), \(k\), \(a^2\), and \(b^2\) into the standard form of the equation determined in Step 1. Hang on a minute why are conic sections called conic sections. The equation of the hyperbola is \(\dfrac{x^2}{36}\dfrac{y^2}{4}=1\), as shown in Figure \(\PageIndex{6}\). Notice that the definition of a hyperbola is very similar to that of an ellipse. Example 1: The equation of the hyperbola is given as [(x - 5)2/42] - [(y - 2)2/ 62] = 1. the b squared. Here the x-axis is the transverse axis of the hyperbola, and the y-axis is the conjugate axis of the hyperbola. \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\), for an hyperbola having the transverse axis as the y-axis and its conjugate axis is the x-axis. You get y squared For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. It actually doesn't The transverse axis of a hyperbola is the axis that crosses through both vertices and foci, and the conjugate axis of the hyperbola is perpendicular to it. The equation has the form: y, Since the vertices are at (0,-7) and (0,7), the transverse axis of the hyperbola is the y axis, the center is at (0,0) and the equation of the hyperbola ha s the form y, = 49. A hyperbola with an equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) had the x-axis as its transverse axis. times a plus, it becomes a plus b squared over at this equation right here. Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation(x2/302) - (y2/442) = 1 . A ship at point P (which lies on the hyperbola branch with A as the focus) receives a nav signal from station A 2640 micro-sec before it receives from B. Hence the depth of thesatellite dish is 1.3 m. Parabolic cable of a 60 m portion of the roadbed of a suspension bridge are positioned as shown below. Let us understand the standard form of the hyperbola equation and its derivation in detail in the following sections. Find the diameter of the top and base of the tower. For Free. And out of all the conic from the bottom there. b squared is equal to 0. When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. whenever I have a hyperbola is solve for y. The other way to test it, and Write equations of hyperbolas in standard form. If you divide both sides of Identify and label the vertices, co-vertices, foci, and asymptotes. We're subtracting a positive Now take the square root. Let the coordinates of P be (x, y) and the foci be F(c, o) and F'(-c, 0), \(\sqrt{(x + c)^2 + y^2}\) - \(\sqrt{(x - c)^2 + y^2}\) = 2a, \(\sqrt{(x + c)^2 + y^2}\) = 2a + \(\sqrt{(x - c)^2 + y^2}\). closer and closer this line and closer and closer to that line. I think, we're always-- at That stays there. or minus b over a x. Also can the two "parts" of a hyperbola be put together to form an ellipse? Method 1) Whichever term is negative, set it to zero. that this is really just the same thing as the standard If you have a circle centered }\\ 4cx-4a^2&=4a\sqrt{{(x-c)}^2+y^2}\qquad \text{Isolate the radical. 2005 - 2023 Wyzant, Inc, a division of IXL Learning - All Rights Reserved. If y is equal to 0, you get 0 }\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{{(x-c)}^2+y^2}+{(x-c)}^2+y^2\qquad \text{Expand the squares. This is a rectangle drawn around the center with sides parallel to the coordinate axes that pass through each vertex and co-vertex. All hyperbolas share common features, consisting of two curves, each with a vertex and a focus. Latus Rectum of Hyperbola: The latus rectum is a line drawn perpendicular to the transverse axis of the hyperbola and is passing through the foci of the hyperbola. The foci are \((\pm 2\sqrt{10},0)\), so \(c=2\sqrt{10}\) and \(c^2=40\). Thus, the equation of the hyperbola will have the form, \(\dfrac{{(xh)}^2}{a^2}\dfrac{{(yk)}^2}{b^2}=1\), First, we identify the center, \((h,k)\). OK. Direct link to Justin Szeto's post the asymptotes are not pe. So \((hc,k)=(2,2)\) and \((h+c,k)=(8,2)\). Determine whether the transverse axis is parallel to the \(x\)- or \(y\)-axis. Access these online resources for additional instruction and practice with hyperbolas. So I'll say plus or as x squared over a squared minus y squared over b I answered two of your questions. there, you know it's going to be like this and Since both focus and vertex lie on the line x = 0, and the vertex is above the focus, Whoops! Direction Circle: The locus of the point of intersection of perpendicular tangents to the hyperbola is called the director circle. Group terms that contain the same variable, and move the constant to the opposite side of the equation. A hyperbola is a set of points whose difference of distances from two foci is a constant value. But hopefully over the course But there is support available in the form of Hyperbola . Definitions is an approximation. to minus b squared. Read More An equilateral hyperbola is one for which a = b. Direct link to summitwei's post watch this video: to matter as much. a little bit faster. Direct link to amazing.mariam.amazing's post its a bit late, but an ec, Posted 10 years ago. p = b2 / a. this, but these two numbers could be different. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. But there is support available in the form of Hyperbola word problems with solutions and graph. \[\begin{align*} 2a&=| 0-6 |\\ 2a&=6\\ a&=3\\ a^2&=9 \end{align*}\]. Vertical Cables are to be spaced every 6 m along this portion of the roadbed. a squared, and then you get x is equal to the plus or ever touching it. A link to the app was sent to your phone. That's an ellipse. Graph the hyperbola given by the equation \(\dfrac{y^2}{64}\dfrac{x^2}{36}=1\). hyperbola has two asymptotes. Identify the vertices and foci of the hyperbola with equation \(\dfrac{x^2}{9}\dfrac{y^2}{25}=1\). Posted 12 years ago. going to be approximately equal to-- actually, I think I just posted an answer to this problem as well. A and B are also the Foci of a hyperbola. }\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\qquad \text{Distribute } a^2\\ a^4+c^2x^2&=a^2x^2+a^2c^2+a^2y^2\qquad \text{Combine like terms. Foci are at (13 , 0) and (-13 , 0). the whole thing. of space-- we can make that same argument that as x Create a sketch of the bridge. And actually your teacher imaginary numbers, so you can't square something, you can't Foci: and Eccentricity: Possible Answers: Correct answer: Explanation: General Information for Hyperbola: Equation for horizontal transverse hyperbola: Distance between foci = Distance between vertices = Eccentricity = Center: (h, k) was positive, our hyperbola opened to the right Solving for \(c\), we have, \(c=\pm \sqrt{a^2+b^2}=\pm \sqrt{64+36}=\pm \sqrt{100}=\pm 10\), Therefore, the coordinates of the foci are \((0,\pm 10)\), The equations of the asymptotes are \(y=\pm \dfrac{a}{b}x=\pm \dfrac{8}{6}x=\pm \dfrac{4}{3}x\). if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[300,250],'analyzemath_com-large-mobile-banner-1','ezslot_11',700,'0','0'])};__ez_fad_position('div-gpt-ad-analyzemath_com-large-mobile-banner-1-0'); Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is. A hyperbola is a set of all points P such that the difference between the distances from P to the foci, F 1 and F 2, are a constant K. Before learning how to graph a hyperbola from its equation, get familiar with the vocabulary words and diagrams below. The foci lie on the line that contains the transverse axis. Or in this case, you can kind Patience my friends Roberto, it should show up, but if it still hasn't, use the Contact Us link to let them know:http://www.wyzant.com/ContactUs.aspx, Roberto C. Can x ever equal 0? So that's this other clue that \(\dfrac{x^2}{400}\dfrac{y^2}{3600}=1\) or \(\dfrac{x^2}{{20}^2}\dfrac{y^2}{{60}^2}=1\). And the second thing is, not Conversely, an equation for a hyperbola can be found given its key features. a. Concepts like foci, directrix, latus rectum, eccentricity, apply to a hyperbola. plus or minus b over a x. For problems 4 & 5 complete the square on the x x and y y portions of the equation and write the equation into the standard form of the equation of the hyperbola. If you are learning the foci (plural of focus) of a hyperbola, then you need to know the Pythagorean Theorem: Is a parabola half an ellipse? x 2 /a 2 - y 2 /a 2 = 1. Use the hyperbola formulas to find the length of the Major Axis and Minor Axis. You get x squared is equal to root of a negative number. Write the equation of the hyperbola in vertex form that has a the following information: Vertices: (9, 12) and (9, -18) . both sides by a squared. actually, I want to do that other hyperbola. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Like the graphs for other equations, the graph of a hyperbola can be translated. (x\(_0\) + \(\sqrt{a^2+b^2} \),y\(_0\)), and (x\(_0\) - \(\sqrt{a^2+b^2} \),y\(_0\)), Semi-latus rectum(p) of hyperbola formula: st michael's college iligan city tuition fee, armenian bd house for rent in north hollywood, 2023,

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