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In triangle ABC, a = 4 and b = c = 2 sqr...

In triangle `ABC, a = 4` and `b = c = 2 sqrt(2)`. A point P moves within the triangle such that the square of its distance from BC is half the area of rectangle contained by its distance from the other two sides. If D be the centre of locus of P, then

A

locus of P is an ellipse with eccentricity `sqrt((2)/(3))`

B

locus of P is a hyperbola with eccentricity `sqrt((3)/(2))`

C

area of the quadr5ilateral `ABCD = (16)/(3)` sq. units

D

area of the quadrilateral `ABCD = (32)/(3)` sq. units

Text Solution

Verified by Experts

The correct Answer is:
A, C

`PM = k`
Equation of `AB -= x +y =2`
Equation of `AC -= x +y =2`
According to question
`((2-h-k)/(sqrt(2)))((2+h+k)/(sqrt(2))) =2k^(2)`
`rArr h^(2) + 3k^(2) + 4k = 4`
`rArr h^(2) + 3 (k^(2)+(4)/(3)k+(4)/(9)) = 4+(4)/(3)`
`rArr h^(2) + 3 (k+(2)/(3))^(2) = (16)/(3) rArr (h^(2))/(16//3) + ((k+(2)/(3))^(2))/(16//9) =1`
`rArr` Ellipse with `e = sqrt((2)/(3))` and `D -= (0,-(2)/(3))`.
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Knowledge Check

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