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  • Find the equation of the hyperbola with (i) Vertices (± 5, 0), foci (± 7, 0)(ii) Vertices (0, ± 7), e = (iii) Foci (0,±√10 ) , passing through (2, 3)

Find the equation of the hyperbola with

(i) Vertices (± 5, 0), foci (± 7, 0)

(ii) Vertices (0, ± 7), e = \frac{4}{3}

(iii) Foci \left ( 0,\pm \sqrt{10} \right ), passing through (2, 3)

 

Answers (1)

a)  Given that a=5 and ae=7

e=7/5

b2=a2(e2-1)=25(49/25-1)=24

So, the equation of hyperbola is 

\frac{x^{2}}{25}-\frac{y^{2}}{24}=1

b) b=7,e=4/3

a^{2}=b^{2}\left ( e^{2} - 1\right )=\frac{343}{9}

\frac{x^{2}}{\frac{343}{9}}-\frac{y^{2}}{49}=-1

9x^{2}-7y^{2}+343=0

c) Given that foci=(0,±10 ) be= \sqrt{10}

a2=b2(e2-1)

a2=b2e2-b2

a2=10-b2

\frac{x^{2}}{10-b^{2}}-\frac{y^{2}}{b^{2}}=-1

Since, hyperbola passes through the point (2,3)

4b2-9(10-b2)=-b2(10-b2)

b4-23b2+90=0

b4-18b2-5b2+90=0

(b2-18)(b2-5)=0

b2=5

a2+b2=10⇒a2=5

\frac{x^{2}}{5}-\frac{y^{2}}{5}=-1

y2-x2=5

 

Posted by

infoexpert22

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