107. Equation of the hyperbola whose vertices are (33,0) and foci at (5,0), is (a) 16x2 - 9y2 = 144 (b)9x2 + 16y2 = 144 (c) 25x2 + 9y2 = 225 (d) 9x2 + 25y2 = 81

Equation of the hyperbola whose vertices are (+-3,0) and foci at (+-5,0) is a. 16 x^2-9y^2=144 b. 9x^2-16 y^2=144 c. 25 x^2-9y^2=225 d. 9x^2-25 y^2=81

Equation of het hyperbola whose vertices are (+-3,0) and foci at (+-5,0) is 16x^(2)-9y^(2)=144 b.9x^(2)-16y^(2)=144c25x^(2)-9y^(2)=225d.9x^(2)-25y^(2)=81

The equation of the hyperbola whose eccentricity 2 and foci are the foci of the ellips x^(2)//25 +y^(2) //9 =1 is

The foci of the hyperbola 9x^2-16 y^2=144 are

For the hyperbola 9x^2-16y^2=144 .find the vertices, foci and eccentricity

The equation of the ellipse whose vertices are (9,2),(-1,2) and the distance between the foci is 8 unit is (A)9(x-3)^(2)+25(y-4)^(2)=225(B)9(x+3)^(2)+25(y+4)^(2)=225(C)9(x-4)^(2)+25(y-3)^(2)=225(D)9(x+4)^(2)+25(y+3)^(2)=225

The equation of the tangent to the hyperbola 16x^(2)-9y^(2)=144 at (5,8//3) , is