1.

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

Answer»


Solution :Tangent at `(sqrt(5),4/3)` is `sqrt(5)x+3y=9x=1` is DIRECTRIX of Hyperbola.
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`


Discussion

No Comment Found

Related InterviewSolutions