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Integral part of the area of figure bounded by the tangents at the end of latus rectum of ellipse (x^(2))/9+(y^(2))/4=1 and directices of hyperbola (x^(2))/9=(y^(2))/72=1is |
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Answer» So, `A=(1,(9-sqrt(5))/3)` & `B=(0,3)` AREA of figure `OGAB` is `1xx((9-sqrt(5))/3)+1/2xx1xx(3-(9-sqrt(5))/3)` `=(9-sqrt(5))/3+(sqrt(5))/6` `=(18-sqrt(5))/6` Area of required figure is `4{(18-sqrt(5))/6}=(36-2sqrt(5))/3` Integral part of area is `7`
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