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Grade: 12th pass


Let ax-y+7=0 be a chord of parabola x^2=28y meeting it at A and B ,and tangents at A and B meet at C.the locus of circumcentre of tr.ABC is

6 years ago

Answers : (1)

Sandeep Pathak
askIITians Faculty
25 Points
							Let us first solve for end points of the chord by finding points of intersection of chord and parabola:
x^2 = 28(ax+7)\\ \Rightarrow x_{\pm} = 14(a\pm\sqrt{a^2+1}), y_{\pm} = \frac{x_{\pm}^2}{28}
It is also interesting to notex_{-}=-x^{-1}_{+}
Let’s now call the slope of tangents at these two points –m_{\pm}
So it is easy to see that
m_+\times m_- = -1

Thus, the two tangents are perpendicular and tr.ABC is a right-angled triangle.

Circumcenter of a right-angled triangle is the midpoint of the hypotenuse.

This gives, for the coordinates of circumcenter,
x_o=14a\\ y_o=\frac{1}{28}\left(x_+^2+x_-^2 \right )=7(2a^2+1)
Thus, the locus is a parabola given in parametric form above.

6 years ago
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