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Grade: 12th pass

                        

From points on the circle x 2 + y 2 = a 2 tangents are drawn to hyperbola x 2 -y 2 =a 2 . Prove that locus of middle points of chords of contact is the curve (x 2 -y 2 ) 2 = a 2 (x 2 +y 2 ).

3 years ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
8397 Points
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5 months ago
Arun
24737 Points
							
Let a point P(x1, y1) be on the circle C1: x²+ y² = a².       --- (1)
Let the tangents to the Hyperbola H1: x² - y² = a²  ---- (2), from P be PQ and PR, touching H1 at Q(x2, y2) and R(x3, y3).
   So PQ:  x *x2 - y * y2 = a², and  PR : x * x3 - y * y3 = a²       --- (3)
 
As PQ and PQ pass through P, 
       x1 * x2 - y1* y2 = a²  and   x1 * x3 - y1 * y3 = a²      --- (4)
Equation of QR - Chord of Contact containing Q & R - is clearly,
           x1 * x - y1 * y = a².    --- (5)
Midpoint of chord of contact QR is: S(α, β) = [ (x2+x3)/2, (y2 +y3)/2 ].
Adding two equations in (4), we get  x1 α - y1 β = a²    --- (6)
Equation of chord of contact QR of H1 with its mid point at S(α, β) is given by formula: 
                                 T    =   S1    --- (7)
            ie., x α - β y - a²  =  α² - β² - a²
           =>  x α - y β = α² - β².      --- (8)
Equations (5) and (8) represent the same Chord of contact QR:
    So   x1 / α = y1 / β = a²/(α² - β²)
     or   x1 = α a²/(α²+β²)   and   y1 = β a²/(α² - β²)      --- (9)
Substitute (9) in eq (1) to get :
         (a⁴ α² + a⁴ β² )/(α² - β²)² = a²
Replace S(α, β) by (x,y) to get the locus.
       =>  (x² + y²) a² = (x² - y²)²
 
5 months ago
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