find equation of hyperbola given foci (0,+_ROOT 10) passing through (2,3)
Answers
Answer:
Step-by-step explanation:
we know that centre C is the midpoint of foci
Then clearly C(0,0)
since the transverse axis is along y-axis,
the equation of the hyperbola is
since it passes through(2,3)
Therefore the equation of the hyperbola is
Answer:
Step-by-step explanation:
\begin{gathered}Given:{F_1}(0,\sqrt{10}), {F_2}(0,-\sqrt{10})\\Distance between foci,{F_1}{F_2}=2\sqrt{10}\\2ae=2\sqrt{10}\\ae=\sqrt{10}\end{gathered}
Given:F
1
(0,
10
),F
2
(0,−
10
)
Distancebetweenfoci,F
1
F
2
=2
10
2ae=2
10
ae=
10
\begin{gathered}b^2=a^2(e^2-1)\\b^2=(ae)^2-a^2\\b^2=10-a^2\end{gathered}
b
2
=a
2
(e
2
−1)
b
2
=(ae)
2
−a
2
b
2
=10−a
2
we know that centre C is the midpoint of foci
Then clearly C(0,0)
since the transverse axis is along y-axis,
the equation of the hyperbola is
\begin{gathered}\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\\\\\frac{y^2}{a^2}-\frac{x^2}{10-a^2}=1\end{gathered}
a
2
y
2
−
b
2
x
2
=1
a
2
y
2
−
10−a
2
x
2
=1
since it passes through(2,3)
\begin{gathered}\frac{9}{a^2}-\frac{4}{10-a^2}=1\\\\\frac{9(10-a^2)-4a^2)}{a^2(10-a^2)}=1\end{gathered}
a
2
9
−
10−a
2
4
=1
a
2
(10−a
2
)
9(10−a
2
)−4a
2
)
=1
\begin{gathered}90-13a^2=10a^2-a^4\\a^4-23a^2+90=0\\(a^-5)(a^2-18)=0\\a^2=5,18\end{gathered}
90−13a
2
=10a
2
−a
4
a
4
−23a
2
+90=0
(a
−
5)(a
2
−18)=0
a
2
=5,18
\begin{gathered}when\:a^2=18, b^2=10-18=-8\\which\:is\:impossible\\\\ when\:a^2=5,\:b^2=5\end{gathered}
whena
2
=18,b
2
=10−18=−8
whichisimpossible
whena
2
=5,b
2
=5
Therefore the equation of the hyperbola is
\begin{gathered}\frac{y^2}{5}-\frac{x^2}{5}=1\\\\y^2-x^2=5\end{gathered}
5
y
2
−
5
x
2
=1
y
2
−x
2
=5