Grade 12th passAnalytical Geometry

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

Gaganpreet Singh

12 Years agoGrade 12th pass

1 Answer

Sandeep Pathak

11 Years ago

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 note $x_{-}=-x^{-1}_{+}$
Let’s now call the slope of tangents at these two points – $m_{\pm}$
$m_{\pm}=\frac{dy}{dx}\Big|_{x_{\pm}}=\frac{x_{\pm}}{14}$
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.
AND

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.