IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

nuksi_nuksi

suppose u are provided with the transverse and conjugate axis of a yperbola as x = y and x+y = 0 and their lenghts are given as 2 and 1 how will we derive the equation of the hyperbola ??

edited previous message .

pls try now

nuksi_nuksi

anyone pls ?

nuksi_nuksi

edited ....... pls try now

nuksi_nuksi

anyone try please

allamraju

I think the ans is (x-y)²/4-(x+y)²/1=1.Is it correct?if so,I will explain.

the above eqn can be written as 3x²+3y²+10xy+4=0

celestine

(x-y)^2 - 4(x+y)^2 =2

3x^2 + 3y^2 + 10xy +2 =0

Im sure tis is correct

allamraju

I think I made a mistake.since 2a and 2b are 2 & 1,a=1 & b=1/2

so,the eqn. is (x-y)²/1-4(x+y)²=1

or 3x²+3y²+10xy+1=0

celestine

hey allam another mistake u mad is that X=(x-y)\root 2 and same for Y also . hence const term is 2

nuksi_nuksi

hye celestine...... yes i think u are correct

please tell me how u approached this question

nuksi_nuksi

hey .. pls reply ..... i mean is it possible to objectively solve this ?

celestine

nuksi

this is concerned with rotation of axes

nuksi_nuksi

hey thanks man ......

celestine

X= (x -y) cosO
Y= (x + y)sinO in new convention
here O= 45*

nuksi_nuksi

hey buddy

i got one more doubt plsss ..

isn;t X = x cosO + y sinO

and Y = y cosO - y sin O

how can we directly substitute the line's equation ...

and also if the line's equation would have been x-y+2 = 0 and x+y =3 then how will we approach further ...

allamraju

Thanks celestine,I made a silly mistake.

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I think the ans is (x-y)2/4-(x+y)2/1=1.Is it correct?if so,I will explain.

the above eqn can be written as 3x2+3y2+10xy+4=0

I think I made a mistake.since 2a and 2b are 2 & 1,a=1 & b=1/2

so,the eqn. is (x-y)2/1-4(x+y)2=1

or 3x2+3y2+10xy+1=0

I think the ans is (x-y)²/4-(x+y)²/1=1.Is it correct?if so,I will explain.

the above eqn can be written as 3x²+3y²+10xy+4=0

so,the eqn. is (x-y)²/1-4(x+y)²=1

or 3x²+3y²+10xy+1=0