Let the equation of two diameters of a circle x2+y2-2 x+2 f y+1=0 be 2 p x-y=1 and 2 x+p y=4 p. Then the slope m ∈(0, ∞) of the tangent to the hyperbola 3 x2-y2=3 passing through the centre of the circle is equal to

Question

Let the equation of two diameters of a circle x2+y2-2 x+2 f y+1=0 be 2 p x-y=1 and 2 x+p y=4 p. Then the slope m ∈(0, ∞) of the tangent to the hyperbola 3 x2-y2=3 passing through the centre of the circle is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 p + f -1=0 .....(1) 2-p f -4 p =0 .....(2) 2=p( f +4) p =(2/ f +4) 2 p =1- f (4/ f +4)=1- f f 2+3 f =0 f =0 text or -3 Hyperbola 3 x2-y2=3, x2-(y2/3)=1 y = mx ± √ m 2-3 It passes (1,0) o = m ± √ m 2-3 m tends ∞ It passes (1,3) 3=m ± √m2-3 (3-m)2=m2-3 m=2

Q. Let the equation of two diameters of a circle x2+y2−2x+2fy+1=0 be 2px−y=1 and 2x+py=4p. Then the slope m∈(0,∞) of the tangent to the hyperbola 3x2−y2=3 passing through the centre of the circle is equal to _____

2p+f−1=0.....(1) 2−pf−4p=0.....(2) 2=p(f+4) p=f+42​ 2p=1−f f+44​=1−f f2+3f=0 f=0 or −3 Hyperbola 3x2−y2=3,x2−3y2​=1 y=mx±m2−3​ It passes (1,0) o=m±m2−3​ m tends ∞ It passes (1,3) 3=m±m2−3​ (3−m)2=m2−3 m=2

Q. Let the equation of two diameters of a circle $x^{2} + y^{2} - 2 x + 2 f y + 1 = 0$ be $2 p x - y = 1$ and $2 x + p y = 4 p$ . Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola $3 x^{2} - y^{2} = 3$ passing through the centre of the circle is equal to _____