If the eccentricity of the hyperbola is 5/4
Web11 sep. 2024 · Answer is (2) 4√5 3 4 5 3 Given θ ∈ (0,π/2) equation of hyperbola ⇒ x2–y2sec2θ = 10 Hence eccentricity of hyperbola Now equation of ellipse ⇒ x2sec2θ + y2 = 5 Hence eccenticity of ellipse Now length of latus rectum of ellipse ← Prev Question Next Question → JEE Main 2024 Test Series NEET Test Series Class 12 Chapterwise MCQ … Web30 jan. 2024 · Equation of the hyperbola whose eccentricity is 5/4 focus (a,0) and directrix 4x-3y=a is 7y²+24xy-24ax-16ay+15a² = 0 Given, the eccentricity is 5/4 Focus is (a,0) Directrix is 4x - 3y - a = 0 So, the equation of the hyperbola would stand as: (x-a)² + (y …
If the eccentricity of the hyperbola is 5/4
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Web5 4 C. If A B is a double ordinate of a hyperbola a 2 x 2 − 9 y 2 = 1 such that O A B is an equilateral triangle of side 2, then eccentricity of hyperbola is equal to (where O is centre of hyperbola) 3. 3 5 D. If the foci of ellipse k 2 a 2 x 2 + a 2 y 2 = 1 and the hyperbola a … Web9 jan. 2024 · Solution: Find the equation of the hyperbola given the asymptotes and passes through a point; Solution: What is the equation of the asymptote of the hyperbola x^2/9 – y^2/4=1? Solution: The semi-transverse axis of the hyperbola x^2/9 – y^2/4 = 1 is? …
Web1. The particular kind of hyperbola in which the lengths of the transverse and conjugate axis are equal is called an equilateral hyperbola. 2. Eccentricity of equilateral hyperbola = √2 3. Equation of pair of asymptotes of rectangular hyperbola x2 − y2 = a2 is x2 − y2 = 0 4. WebThe eccentricity of hyperbola can be found from the formula e = √1 + b2 a2 e = 1 + b 2 a 2. For this formula, the values a, and b are the lengths of semi-major axes and semi-minor axes of the hyperbola. And these values can be calculated from the equation of the …
Web22 mrt. 2024 · Ex 11.4, 1 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola x2 16 - y2 9 = 1 Given equation is 2 16 2 9 = 1 The above equation is of the form 2 2 2 2 = 1 So axis of Web3 jan. 2024 · Find the equation of the parabola with vertex at (4, 3) and focus at (4, -1). A. y 2 – 8x + 16y – 32 = 0 B. y 2 + 8x + 16y – 32 = 0 C. y 2 + 8x – 16y + 32 = 0 D. x 2 – 8x + 16y – 32 = 0 View Answer: 61. Find the area bounded by the curves x 2 + 8y + 16 = 0, x – 4 = 0, the x-axis, and the y-axis. A. 10.67 sq. units B. 10.33 sq. units
Web(Mar-11,17 TS) If the eccentricity of a hyperbola is 5/4, then find the eccentricity of its conjugate hyperbola. (June-05,Mar - 2013) (Mar-15 T.S) (Mar&May-16,17 A.P) See answers Advertisement mohdsaqib179 Answer: eccentricity of the conjugate hyperbola …
WebThe eccentricity of an ellipse which is not a circle is greater than zero but less than 1. The eccentricity of a parabola is 1. How do you find the eccentricity of a rectangular hyperbola? Solution: The eccentricity of the standard equation of the hyperbola x2a2 … oval metal woven patio furnitureWebThe linear eccentricity (focal distance) is c = \sqrt {a^ {2} + b^ {2}} = 3 \sqrt {5} c = a2 + b2 = 3 5. The eccentricity is e = \frac {c} {a} = \frac {\sqrt {5}} {2} e = ac = 25. The first focus is \left (h - c, k\right) = \left (- 3 \sqrt {5}, 0\right) (h − c,k) = (−3 5,0). oval micro pave halo engagement ringWeb1. The particular kind of hyperbola in which the lengths of the transverse and conjugate axis are equal is called an equilateral hyperbola. 2. Eccentricity of equilateral hyperbola = √ 2 3. Equation of pair of asymptotes of rectangular hyperbola x 2 − y 2 = a 2 is x 2 − y 2 = … イチネンケミカルズ ペネトンaWebThe transverse axis of a hyperbola is the line that contains the two vertices and the two focuses. In this example (hyperbola of equation 2x 2− 4y 2=1 ), the transverse axis is the X− Axis graph {x 2/2−y 2/4=1[−10,10,−5.25,5.17]} In this example (hyperbola of equation 2x 2− 4y 2=−1) the transverse axis is the Y− Axis oval mirrored medicine cabinet recessedWebd) 4 x 2 - 5y 2 – x + 8y + 1 = 0. See Summaries on page 588. Eccentricity. A conic is the set of all points P(x, y)in a plane such that the ratio of the distance from P to a fixed point and the distance from P to a fixed line is constant. The constant ratio is called the … oval mirror medicine cabinet recessedWeb7 dec. 2024 · 1. Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity. Asymptote : 5 x − 4 y + 5 = 0 and Tangent : 4 x + 5 y − 7 = 0. I thought that if we consider asymptote to be limiting … oval mirror medicine cabinet vintageWeb7 dec. 2024 · The eccentricity of the above hyperbola is e = 2 m 1 + m 2 − 1 + m 2, hence it can take any value between 1 and 2, depending on the value of m. EDIT. With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation ( 5 x − 4 y + 5) ( ( 4 − 5 m) y − ( 5 + 4 m) x − 5 + 25 4 m) = 9 64 m 2 oval montalban