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If the foci of the ellipse `(x^(2))/(16)+(y^(2))/(b^(2))=1` and the hyperbola `(x^(2))/(144)-(y^(2))/(81)=(1)/(125)` coincide, the find the value of `b^(2)`. |
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Answer» Correct Answer - 7 For a hyperbola `e^(2)=1+(b^(2))/(a^(2))=1+(81)/(144)=(225)/(144)` `:. E = (15)/(12)=(5)/(4)` Also, `a^(2)=(144)/(25)` Hence, the foci are `(+- "ae", 0)-+(+-(12)/(5),(5)/(4))-=(+-3,0)` Now, for an ellipse, ae = 3 or `a^(2)e^(2)=9` Now, `b^(2)=a^(2)(1-e^(2))` `rArr b^(2)=a^(2)-a^(2)e^(2)=16-9=7` |
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