Q: The vertices of the hyperbola 9 x^2-16 y^2-36 x+96 y-252=0 are

Given hyperbola is aligned & 9 x^2-16 y^2-36 x+96 y-252=0 \\ & 9 (x^2-4 x )-16 (y^2-6 y )=252 \\ & 9 (x^2-4 x+4 )-16 (y^2-6 y+9 ) \\ & =252+36-144 \\ & 9(x-2)^2-16(y-3)^2=144 aligned Put x=X+2 and y=Y+3 Equation (i) becomes X^2 16 - Y^2 9 =1 Now, vertices are X= a wehre a=4 and Y=0 Hence, vertices are (6,3),(-2,3).

Question 1

The foci of hyperbola 4 x^2-9 y^2-1=0 are

Accepted Answer

Given, Hyperbola =4 x^2-9 y^2-1=0 aligned & x^2 ( 1 4 ) - y^2 ( 1 9 ) =1 \ & x^2 1 2 ^2 - y^2 1 3 ^2 =1 aligned We know that. if hyperbola is x^2 a^2 - y^2 b^2 =1. then Eccentricity: e= 1+ b^2 a^2 aligned & and focus ( be , 0)= ( 1 2 1 / 3 3 , 0 )= ( 13 6 , 0 ) aligned

Question 2

The vertices of the hyperbola 9 x^2-16 y^2-36 x+96 y-252=0 are

Accepted Answer

Given hyperbola is aligned & 9 x^2-16 y^2-36 x+96 y-252=0 \ & 9 (x^2-4 x )-16 (y^2-6 y )=252 \ & 9 (x^2-4 x+4 )-16 (y^2-6 y+9 ) \ & =252+36-144 \ & 9(x-2)^2-16(y-3)^2=144 aligned Put x=X+2 and y=Y+3 Equation (i) becomes X^2 16 - Y^2 9 =1 Now, vertices are X= a wehre a=4 and Y=0 Hence, vertices are (6,3),(-2,3).

Question 3

If the eccentricity and length of latus rectum of a hyperbola are 13 3 and 10 3 units respectively, then what is the length of the transverse axis?

Accepted Answer

Length of latus rectum of a hyperbola is 2 b^2 a where a is the half of the distance between two vertex of the hyperbola. Latus rectum = 2 b^2 a = 10 3 or, b^2= 5 a 3 ...(i) In case of hyperbola, b ^2= a ^2 ( e ^2-1 ) ...(ii) Putting value of b^2 from equation (i) and e= 13 3 in equation (ii), 5 a 3 = a ^2 ( 13 9 -1 ) or, 5 a 3 = 4 a^2 9 4 a^2-15 a=0 or, a(4-15 a)=0 a 0, hence, a= 15 4

Hyperbola — Previous Year Questions & Solutions