tis is question fron chapter hyperbola (class11)
the question is "Find the equation of the hyperbola,referred to its axes as the axis of co-ordinate, whose cojugate axis is 5 and which passes through the point(1,-2). Find also the equation of the conjugate hyperbola.
please do it very clearly without any confusion.
and the answer given in my book is
;41x^2-4y^2-25=0 ;4-x^2-4y^2+25=0
2007-01-20 02:00:50 · 3 answers · asked by suchi 2 in Science & Mathematics ➔ Mathematics
The equation of the hyperbola is x^2/a^2-y^2/b^2=1
conjugate axis=5 (given) and so b=5/2 and b^2=25/4
substituting b^2=25/4 in the equation
x^2/a^2-4y^2/25=1
Passes through (1,-2)
So substituting x=1 and y=-2
1/a^2-16/25=1
1/a^2=1+16/25=41/25
a^2=25/41
so the equation is
x^2/(25/41)-y^2/25/4)=1
41x^2/25-4y^2/25=1
=>41x^2-4y^2=25
=>41x^2-4y^2-25=0
The equation of the conjugate hyperbola will be
X^2/a^2-y^2/b^2=-1
So 41x^2/25-4y^2/25=-1
41x^2-4y^2=-25
=>41x^2-4y^2+25=0
2007-01-24 04:40:19 · answer #1 · answered by raj 7 · 1⤊ 0⤋
very gud explaination or answer by raj
2007-01-26 23:19:15 · answer #2 · answered by n nitant 3 · 0⤊ 0⤋
I dont know
2007-01-25 04:04:13 · answer #3 · answered by Brent S 1 · 0⤊ 1⤋