Solving
1. Equation of the tangent to the parabola
The given parabola is y2=6x. Comparing with y2=4ax, we get 4a=6, so a=23.
The equation of a tangent to y2=4ax in slope form (m) is y=mx+ma.
Substituting a:
2. Finding the vertices of the triangle
The triangle is formed by the intersection of three lines:
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Line 1: x=0
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Line 2: y=3
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Line 3 (Tangent): mx−y+2m3=0
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Intersection of x=0 and y=3: Vertex A(0,3).
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Intersection of x=0 and Tangent: y=2m3. Vertex B(0,2m3).
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Intersection of y=3 and Tangent: 3=mx+2m3⇒mx=3−2m3⇒x=2m26m−3. Vertex C(2m26m−3,3).
3. Identifying the Circumcentre
Since Line 1 (x=0, vertical) and Line 2 (y=3, horizontal) are perpendicular, the triangle is a right-angled triangle at vertex A(0,3).
In a right-angled triangle, the circumcentre (h,k) is the midpoint of the hypotenuse (the side opposite the right angle, which is BC).
4. Coordinates of the Circumcentre (h,k)
h=2xB+xC=20+2m26m−3=4m26m−3
k=2yB+yC=22m3+3=4m3+6m
5. Eliminating the parameter m
From the equation for k:
4k=m3+6⇒m3=4k−6⇒m1=34k−6
From the equation for h:
4h=m26m−m23=m6−3(m1)2
Substitute m1=34k−6:
Multiply the whole equation by 3:
6. Final Equation
Divide by 4:
Replacing (h,k) with (x,y):
Correct Option: (3)
