11centre (-1,3)
conjugate axis 2x+y+1=0 (incase it is wrong this x+2y-1=0)
eccentricity ????????
The asymptotes of a hyperbola are x^2+4xy+y^2-2x+2y-2=0 and the hyperbola passes through (1,1)
The centre of the hyperbola is?
The conjugate axis of the hyperbola is?
What is the eccentricity of the hyperbola?
-
UP 0 DOWN 0 0 6
6 Answers
111The options for the second question are
(a)x-y+2=0 (b)x=y (c)x+y=2 (d)x=-y
please provide the full solution
11yeah second one wrong :(
for the first one partial differentiate the equation first wrtx and then
wrty
u get two equations
solve them to get centre
1i worked out the 3rd question,
on transforming the axes,the equation becomes
\frac{x^2}{\frac{4}{3}}-\frac{(y-\sqrt{2})^2}{4}=1
whose e=2
please solve the second question
62you have found the hyperbola... in the standard form
then you can find the conjugate equation... then transform it back to the original form
btw this is not for jee syllabus..
i dont think you should study this if you are giving jeee this year..
If you have just got into xii you should solve it :)