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Grade: 12th pass
        
Please solve hyperbola question given in attachment as soon as possiblr
one month ago

Answers : (1)

Aditya Gupta
1714 Points
							
hello yuvi, this ques is not even of hyperbola!! when you draw the figure of hyperbola and the chord, you will find that the chord intersects the hyperbola in 2nd and 3rd quads. now, alpha^2 + beta^2 less than equal to 1 implies R lies inside the unit circle. obviously P and Q are fixed points. so, area of PQR= ½ *PQ*h, where h is the height of perpendicular dropped from point R onto the chord 2x+y+6=0.
since PQ is fixed, we need to minimise h in order to minimise the area.
if we draw a perpendicular from the origin onto the chord, then the point of intersection of this perpendicular with the circle will clearly be the point R having the least distance from the chord (it is quite obvious once you draw the diagram).
so, the eqn of perpendicular from origin is y= mx, where m*( – 2)= – 1 (since slope of chord is – 2).
or m= ½ 
so 2y= x is the eqn of chord.
now, solving it with x^2+y^2= 1, we get
(x, y)= ( – 2/sqrt5, – 1/sqrt5) or (2/sqrt5,1/sqrt5). but (2/sqrt5,1/sqrt5) is rejected as it lies in the first quad.
so, the coordinates of R:
( – 2/sqrt5, – 1/sqrt5)
option B
kindly approve :)
one month ago
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