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PQ is the double ordinate of the parabola y^(2)=4x which passes through the focus S. DeltaPQA is an isosceles right angle triangle, where A is on the axis of the parabola to the right of focus. Line PA meets the parabola at C and QA meets the parabola at B. The of trapezium PBCQ is |
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Answer» 96 sq. UNITS For `y^(2)=4X`, the coordinates of the ends of latus rectum are P(1,2) and Q(1,-2). `DeltaPAQ` is isosceles RIGHT - angled. Therefore, the slope of PA is -1 and its equation is y-2=-(x-1), i.e., x+y-3=0. Similarly, the equation of line QB is x-y-3=0. Solving x+y-3=0 with the parabola `y^(2)=4x`, we have `(3-x)^(2)=4x,i.e.,x^(2)-10x+9=0` `:.x=1,9` Therefore, the coordinates of B and C are (9,-6) and (9,6), respectively. Area of trapezium `PBCQ=(1)/(2)xx(12+4)xx8=64` sq. units Let the circumcenter of trapezium PBCQ be T(h,0). Then, PT=BT `orsqrt((h-1)^(2)+4)=sqrt((h-9)^(2)+36)` `or-2h+=-18h+81+36` `or16h=112` `orh=7` Hence, radius is `sqrt(40)=2sqrt(10`. Let the INRADIUS of `DeltaAPQ` be `r_(1)`. Then, `r_(1)=(Delta_(1))/(s_(1))` `((1)/(2)xx4xx2)/((1)/(2)(4+2sqrt(4+4)))` `=(2)/(1+sqrt(2))=2(sqrt(2)-1)` Let the inradius of `DeltaABC` be `r_(2)`. Then, `r_(2)=(Delta_(2))/(s_(2))` `=((1)/2xx12xx6)/((1)/(2)(12+2sqrt(36+36)))` `=(6)/(1+sqrt(2))=6(sqrt(2)-1)` `:." "(r_(2))/(r_(1))=3` |
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