Question

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Find an equation of the hyperbola having foci at (3.3) and (3.9) and vertices at (3, 5) and (3.7). Ole X $ ? Check © 2020 McGraw-

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Answer 1

Solution : © 5 Given that a hyperbola with foci at (3, 3) & (3, 4) and vertices at (3,5) and (3,7) (29) hele know that the equation of the .(3,9) (3,7) hyperbola is o (4-h? (y-K)² 1 major axis along x-axis az 62 or parallel where (hik) is the centre of the hyperbola Here (h, k) = (3,6) a: distance of the veretter from centre Here a= ci distance of the focus from centre Here e=3 382 7 2 2 762 and b2 = (²_q² = 3 - 1² = 9-) = 8 7 b = 88 = 282 where b is the height of the fundamental box- the major axis is at (n-h² Parallel to Since atting y-axis, equation will be (M-K² 1 62 a ? (4-6) 2. M-3)? 1 8 CS Scanned with CamScanner

y = - = x Since 33 :) (-13,0) (13,0) Given that foci at (-13,0) and (13,0) asymptotes are ya 15 2 and Here centre of the hyperbola at (0,0) y = 1 5 2 n are asymptotes, passing through origin ya 36 y-k = = b (n-h) 7 y = K + b cu-h) ty thax (chik) = (0,0)) 7 a= la and b=5 then it's equation will be and again x y = -5 g - ² = 1 (M-62 az N 62 2 n y2-1 144 25 CS Scanned with CamScannet

of length 1 JO axis=10 y length of semiminar (712 & Co-ordinate of the co-verter Đ Given that center at (012) clinor axis 10 verster at (712) Sinu to length of misre 7(0,2) 10,-3) in the uppen half plane is (0,1) and in the lower half plane it is. (0, -3) the width of Ellipse is from the above diagram, we get 2a = 14 => a = =26=10 ) b = 5 exis = 10 -5 2 height=26=10 .. Equation will be (n-0² 6-5) 1 + 12 4-5)2 - 1 W + 49 CS Scanned with Can Scanner

Y 8 = 12 Given that majon avis length= and for at (-2, 8) and (-2,-2) det F, = (-2,8) Chikta) (478) Fz=(-2,-2) 0 Ae (2-24 V Chik-a) 8 and consider (hik) be the centre of the ellipse then (bi Кt c) = Fi and (h, K-c) = f2 i K+) = (-2,8) CPK-u) = (-2,-2) =) k to ş-02-2 7 2K = 6 77 K-3 and c=5 :: Center of the ellipse is (-2,3) Length of the major anis = 12 729=1279= 6 polypepad Let the vertices be (h, k+a), 2(R, x-a) & Again we know that (² - 2²-6² 62 = 9²-12 6² = 36-25 = 1 (+2) y b =VII =1 Equation is (4-33 + 36 CS Scanned with Can Scanner