Please solve hyperbola question given in attachment as soon as possiblr

Question

Aditya Gupta · Answer

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 2 nd and 3 rd 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 :)

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Please solve hyperbola question given in attachment as soon as possiblr

Please solve hyperbola question given in attachment as soon as possiblr

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