the line x-y+3=0 touches the hyperbola whose foci are (⁺₋√41,0).find the equation of the hyperbola
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Answered by
0
Answer:
Since, the foci of the given hyperbola are F
′
(0,−
10
) and F(0,
10
) which lie on the X-axis and mid-point of the segment F
′
F is (0,0). Therefore origin is the centre of the hyperbola and its transverse axis lies along Y-axis, hence the equation of the hyperbola can be taken as
a
2
y
2
−
b
2
x
2
=1 .........(i)
As the foci are (0,±
10
)
So, ae=
10
We know that
b
2
=a
2
e
2
−a
2
⇒b
2
=10−a
2
Answered by
4
x 2 −2y 2=18
x+y=a
y=(x−a)
⇒x 2 −2(x−a) 2 =18
⇒−x +4ax−2a 2=18
⇒x 2−4ax+2(a 2−18)=0
Since line touches the hyperbola. So only one value of x. Hence D=0
⇒16a 2 −8(a 2 −18)=0
a 2 =9
∣a∣=3.
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