1.

P is a variable point on the line L=0 . Tangents are drawn to the circles x^(2)+y^(2)=4 from P to touch it at Q and R. The parallelogram PQSR is completed.If P-= (6,8), then the area of Delta QRS is

Answer»

`(3sqrt(6))/(25)` sq. units
`(3sqrt(24))/(25)` sq. units
`(48sqrt(6))/(25)` sq. units
`(192sqrt(6))/(25)` sq. units

Solution :`P -= (6,8)`
Thereforem the equation of QR (chord of contact ) is
`6x+8y=4`
or `3x+4y-2=0 `
`:. PM =(48)/(50` and `PQ =SQRT(96)`
`QM =sqrt(96-(48^(2))/(25))=sqrt((96)/(25))`
`:. QR = 2 sqrt((96)/(25))`
`:. ` Area of `Delta PQR =(1)/(2) XX PM xx QR =(192 sqrt(6))/(25)`
PQRS is a rhombus. Therefore,
Area of `DeltaRS =` Area of `Delta PQR = (192sqrt(6))/(25)`


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