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So Question is

The tangent at any point $P$ on the parabola $y^2=4ax$ intersects the $y$-axis at $Q$. Then tangent to the circumcircle of $\triangle PQS$ ($S$ is the focus) at $Q$ is:

(A) a line parallel to $x$-axis (B) $y$-axis
(C) a line parallel to $y$-axis (D) none of these

I assumed point $P$ as $(at^2,2at)$ then from there I got equation of tangent to parabola later I got point $Q$ as $(0,at)$. After having $P,Q,S$ points I am not understanding how to find tangent at $Q$.

Circumcenter I got is $(a(t^2+1)/2, at)$

Mr. K
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  • Please edit the question to show us your working. That way we can help better. What are the coordinates of $P, Q, S$ you obtained? – Math Lover May 22 '21 at 06:23
  • @MathLover The OP has already mentioned the coordinates in parametric form. $S$ is the focus which has the coordinates $(a,0)$. – Shubham Johri May 22 '21 at 06:24
  • To find the tangent at $Q$ you need the slope of the tangent. A straightforward approach is: the tangent is perpendicular to the radial line joining the circumcentre and $Q$, so you need to find the coordinates of the circumcentre. You may already know a formula to calculate that or alternatively find it as the intersection of perpendicular bisectors of $PS$ and $QS$. – Shubham Johri May 22 '21 at 06:25
  • I used the approach that from parametric coordinates P,Q,S I could get equation of circle as there are on the circle and from there equation of tangent to the circle at Q. But the calculation is becoming too long and quite hard with many variables. – Mr. K May 22 '21 at 06:40
  • What is the centre of the circumcircle you got? – Shubham Johri May 22 '21 at 06:44
  • Circumcenter I got is [a(t^2+1)/2 , at] – Mr. K May 22 '21 at 07:15
  • So if you know the coordinates of $O$ (circumcenter), you know the slope of $OQ$. Then slope of tangent to circle is perpendicular to that and you can find that using formula $m_1 m_2 = -1$. So finally you have a point on the tangent line (point $Q$) and its slope. You can find the equation of the tangent line, right? – Math Lover May 22 '21 at 07:24
  • @ShubhamJohri reading your earlier comment, it is unnecessary to use any formula required to find the circumcenter here or to find intersection of perpendicular bisectors. By property of parabola $y^2 = 4ax$, $\angle PQS$ will be $90^0$ and diameter of the circumcircle will be on $PS$ and is simply the midpoint of $PS$. – Math Lover May 22 '21 at 08:06
  • @MathLover could you please tell me the property of the parabola that you have mentioned. – Mr. K May 22 '21 at 08:37
  • For a parabola of the form $y^2 = ax$, the perpendicular from the focus to the tangent line to any point on the parabola will meet at a point on y-axis. The same holds true for parabola of the form $x^2 = ay$ that they meet at x-axis. With the coordinates of vertices you got here, you can check that the product of the slope of $PQ$ and of $QS$ is $-1$, i.e. they are perpendicular. For a right angled triangle, the circumcenter is the midpoint of hypotenuse. So you can very easily obtain the coordinates of circumcenter. – Math Lover May 22 '21 at 08:52
  • https://math.stackexchange.com/questions/665837/prove-that-the-foot-of-the-perpendicular-from-the-focus-to-any-tangent-of-a-para – Intelligenti pauca May 22 '21 at 14:43

1 Answers1

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Slope of the radius joining the circumcentre and $Q$ is $$m=\frac{at-at}{a(t^2+1)-0}=0$$The tangent at $Q$ is perpendicular to this radius, and hence has the slope $\infty$, i.e. the tangent is parallel to the $y$ axis. But the tangent passes through $Q$ which itself lies on $y$-axis, and thus the tangent is the $y$-axis.

Shubham Johri
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