WebMar 30, 2024 · Hence, the required equation of the hyperbola is 𝒙𝟐/𝒂𝟐 – 𝒚𝟐/𝒃𝟐 = 1 Now, coordinates of foci are (±c, 0) & given foci = (±4, 0) so, (±c,0) = (±4,0) c = 4 Now, Latus rectum =2𝑏2/𝑎 Given latus rectum = 12 So, 2𝑏2/𝑎=12 2b2 = 12a b2 = 6a We know that c2 = a2 + b2 Putting value of c & b2 (4)2 = a2 + 6a 16 = a2 + 6a a2 + 6a − 16 = 0 a2 + 8a − 2a −16 = 0 … WebSolution Foci (±4, 0), the latus rectum is of length 12. Here, the foci are on the x -axis. Therefore, the equation of the hyperbola is of the form. Since the foci are (±4, 0), c = 4. Length of latus rectum = 12 We know that a2 + b2 = c2. ∴ a2 + 6 a = 16 ⇒ a2 + 6 a – 16 = 0 ⇒ a2 + 8 a – 2 a – 16 = 0 ⇒ ( a + 8) ( a – 2) = 0 ⇒ a = –8, 2
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WebTherefore, the length of the latus rectum of an ellipse is given as: = 2b 2 /a = 2 (2) 2 /3 = 2 (4)/3 = 8/3 Hence, the length of the latus rectum of ellipse is 8/3. For more Maths-related articles and solved problems, register with BYJU’S – The Learning App and download the app to learn with ease. Quiz on Latus rectum Start Quiz WebAug 19, 2024 · Find the equation of the ellipse in each of the cases given below: (i) foci (± 3, 0), e = 1/2. ... (iii) length of latus rectum 8, eccentricity = 3/5 and major axis on x -axis. (iv) length of latus rectum 4, distance between foci 4√2 and major axis as y -axis. two dimensional analytical geometry; saber tactical stock
Ex 11.4, 10 - Find hyperbola: foci (5, 0), transverse 8 - Ex …
WebMar 9, 2024 · Length of the latus rectum: Length of the latus rectum = 2a 2 /b (when a 2 < b 2) = 2×4/5 = 8/5 Question 3. = 1 Solution: Since denominator of x 2 /16 is larger than the denominator of y 2 /9, the major axis is along the x-axis. Comparing the given equation with = 1, we get a 2 = 16 and b 2 = 9 ⇒ a = ±4 and b = ±3 The Foci: WebMar 16, 2024 · Transcript. Ex 11.4, 7 Find the equation of the hyperbola satisfying the given conditions: Vertices (±2, 0), foci (±3, 0) Given Vertices are (±2, 0) Hence, vertices are on the x-axis ∴ Equation of hyperbola is of the form 𝒙𝟐/𝒂𝟐 – 𝒚𝟐/𝒃𝟐 = 1 Now, Co-ordinate of vertices = (±a, 0) & Vertices = (±2, 0) ∴ (±a, 0 ... WebThis calculator will find either the equation of the hyperbola from the given parameters or the center, foci, vertices, co-vertices, (semi)major axis length, (semi)minor axis length, … saber technologies private limited