Conics nd poly. (1 Viewer)

[chousta]

hi

Q.

A circle passing through the origin is tangent to the hyperbola xy=1 at A, and intersects the hyperbola again at two distinct points B and C.
Prove that OA is (PERPENDICULAR) to BC


cheers:wave:

 

[Yip]

Let the circle equation be (x-a)²+(y-b)²=r²,
Since it passes through the origin, a²+b²=r²
xy=1
y=1/x
Substitute into circle,
(x-a)²+[(1/x)-b]²=r²=a²+b²
x⁴-2ax³-2bx+1=0
Let A(gamma,1/gamma), B(alpha,1/alpha), C(beta,1/beta), where alpha, beta and gamma are the roots of the quartic equation above.
Since the circle is tangential to the hyperbola, and also intersects the hyperbola at 2 other distinct points, a double root exists at A.
Gradient of BC=(1/alpha-1/beta)/(alpha-beta)
=-1/alpha.beta
Gradient of OA=(1/gamma)/gamma
=1/gamma^2
Gradient of BC.Gradient of OA=-1/alpha.beta.gamma.gamma
=-1 (since gamma is a double root, this is simply the product of the roots of the quartic, which can be found by using the root coefficient relations)

 

[chousta]

Let the circle equation be (x-a)²+(y-b)²=r²,
Since it passes through the origin, a²+b²=r²
xy=1
y=1/x
Substitute into circle,
(x-a)²+[(1/x)-b]²=r²=a²+b²
x⁴-2ax³-2bx+1=0
Let A(gamma,1/gamma), B(alpha,1/alpha), C(beta,1/beta), where alpha, beta and gamma are the roots of the quartic equation above.
Since the circle is tangential to the hyperbola, and also intersects the hyperbola at 2 other distinct points, a double root exists at A.
Gradient of BC=(1/alpha-1/beta)/(alpha-beta)
=-1/alpha.beta
Gradient of OA=(1/gamma)/gamma
=1/gamma^2
Gradient of BC.Gradient of OA=-1/alpha.beta.gamma.gamma
=-1 (since gamma is a double root, this is simply the product of the roots of the quartic, which can be found by using the root coefficient relations)

thank tim, much appreciated.