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Distance between the foci of the curve r...

Distance between the foci of the curve represented by the equation `x=3+4cosθ,y=2+3sinθ` is

Text Solution

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(a) The locus is `(x^(2))/(16)+(y^(2))/(36)=1`
`e=sqrt(1-(16)/(36))=sqrt((20)/(36))=(sqrt(5))/(3)`
or `3e=sqrt(5)`
(b) `(3x^(2)-2x+1)2(y^(2)-2y+1)=3+2+1` ltbgt or `((x+1)^(2))/(2)+((y-1)^(2))/(3)=1`
or `e=sqrt(1-(2)/(3))=(1)/(sqrt(3))`
`:. a=sqrt(3),b=sqrt(2)`
`:. "Area"=(1)/(2)xx2sqrt(3)xxsqrt(2)=sqrt(16)`
(c) Eliminnating `theta` from `x=1+4cos theta and y=2+3 sin theta` we have `((x-1)^(2))/(16)+((y-2)^(2))/(9)=1`
Hence a=4 and is `sqrt(17)//4`. Thereforr,
`ae=sqrt(7)`
Hence the distance between the focis is `2//sqrt(7)`
(d) `(x^(2))/(16)+(y^(2))/(7)=1`
`e=sqrt(1-(7)/(16))=(3)/(4)`

One end of the latus rectum is
`(ae,(b^(2))/(a))=(3,(7)/(4))`
Therefore, the equation of teangent is `(3x)/(16+(7)/(4)(y)/(7)=1`
It means the x-axis at (16/3,0) and the y-axis at (0,4).
Hence, then area of quadriilateral is
`2xx(16)/(3)xx4=(128)/(3)`
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