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Question

Find the equation of the hyperbola whose
(i) foci are (6, 4) and (−4, 4) and eccentricity is 2.
(ii) vertices are (−8, −1) and (16, −1) and focus is (17, −1)
(iii) foci are (4, 2) and (8, 2) and eccentricity is 2.
(iv) vertices are at (0 ± 7) and foci at 0,±283.

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Solution

(i) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are 6-42,4+42, i.e. 1, 4.
Let 2a and 2b be the length of the transverse and conjugate axis and let e be the eccentricity.

x-12a2-y-42b2=1
Distance between the two focii = 2ae


2ae=6+42+4-422ae=10ae=5a=52

Also, b2=ae2-a2b2=25-254b2=754

Equation of the hyperbola is given below:
4x-1225-4y-4275=14x2-2x+125-y2-8y+1675=14x2-8x+425-4y2-32y+6475=134x2-8x+4-4y2-32y+64=7512x2-24x+12-4y2+32y-64=7512x2-24x-4y2+32y-127=0

(ii) The centre of the hyperbola is given below:
-8+162,-1-12=4,-1
If the other focus is S'm,n, then it is calculated in the following way:
4=17+m2m=-9 And -1=-1+n2n=-1
Thus, the other focus is -9,-1.

Distance between the vertices: 2a=16+82+-1+122a=24a=12

Distance between the foci:
2c=17+92+-1+122c=26c=13

Also, c2=a2+b2b2=169-144b2=25

Equation of the hyperbola is given below:
x-42144-y+1225=1x2-8x+16144-y2+2y+125=125x2-200x+400-144y2+288y+144=360025x2-200x-144y2-288y-3344=0

(iii) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are 4+82,2+22, i.e. 6, 2.
Let 2a and 2b be the length of the transverse and conjugate axis. Let e be the eccentricity.

x-62a2-y-22b2=1
Distance between the two focii = 2ae


2ae=4-82+2-222ae=16ae=8a=4

Also, b2=ae2-a2b2=64-16b2=48

Equation of the hyperbola:
x-6216-y-2248=1x2-12x+3616-y2-4y+448=13x2-12x+36-y2-4y+4=483x2-36x+108-y2+4y-4=483x2-36x-y2+4y+56=0


(iv) The Vertices of the hyperbola are 0,±7.
b=7
The foci is 0,±283.
be=283


Also, a2=b2e2-1a2=2832-49a2=3439
Therefore, the equation of the hyperbola is-9x2343+y249=1.

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