# Examples on Tangents to Parabolas Set 1

**Example - 17**

Prove that the orthocentre of any triangle formed by three tangents to a parabola lies on its directrix.

**Solution:** Let the equation of the parabola be \({y^2} = 4ax\) so that its directrix is *x + a = 0*. We need to show that the *x*-coordinate of the orthocentre is –a.

Assume any three points on the parabola as \({t_1},{\rm{ }}{t_2},{\rm{ }}{t_3}.\) The points of intersections of the three tangents to the parabola on these three points will be

\[P(a{t_1}{t_2},a({t_1} + {t_2})),\,\,Q(a{t_2}{t_3},a({t_2} + {t_3})),\,\,R(a{t_3}{t_1},a({t_3} + {t_1}))\]

Let us find the equation of the altitude through *P* on *QR*.

The slope of *QR *is \(\begin{align}\frac{{a({t_2} + {t_3}) - a({t_3} + {t_1})}}{{a{t_2}{t_3} - a{t_3}{t_1}}} = \frac{1}{{{t_3}}}\end{align}\)

Therefore, the equation of the altitude through *P* is

\[\begin{align}&\qquad \;\;y - a({t_1} + {t_2}) = - {t_3}(x - a{t_1}{t_2})\\\\&\Rightarrow \quad y + {t_3}x = a({t_1} + {t_2} + {t_1}{t_2}{t_3}) \qquad\qquad \dots \left( 1 \right)\end{align}\]

By symmetry we can write the altitude from *Q* onto *PR* as

\[y + {t_1}x = a({t_2} + {t_3} + {t_1}{t_2}{t_3}) \qquad\qquad\qquad \dots \left( 2 \right)\]

By (1) – (2), we have

\[\begin{align}&\qquad \;\;({t_3} - {t_1})x = a({t_1} - {t_3})\\\\&\Rightarrow \quad x = - a\end{align}\]

Thus, the stated assertion is true.

**Example - 18**

Let *P* be any point on the parabola with focus *F*. A perpendicular *PM* is dropped from *P* onto the directrix as shown :

Prove that the tangent at *P* to the parabola bisects the angle *FPM*.

**Solution:** Let the equation of the parabola be \({y^2} = 4ax,\) and let *P* be the point t. Thus, *F* is the point (a, 0) while N is (–a, 0). By definition, we have

\[PF = PM\]

But *PM* is *PX + XM* i.e., \(a{t^2} + a.\) Thus *PF* = \(a{\rm{ }} + {\rm{ }}a{t^2}\).

Now, the equation of the tangent at *P* is

\[ty = x + a{t^2}\]

This intersects the x-axis at the point \(Q( - a{t^2},0).\)

Thus,

\[\begin{align}&FQ = FO + OQ\\&\quad\;\;= a + a{t^2}\end{align}\]

We see that in \(\Delta PFQ,\)

\[\begin{align}&\qquad \quad FP = FQ \\\\& \Rightarrow \quad \angle FPQ = \angle FQP \qquad \quad \dots \left( 1 \right)\end{align}\]

But since *PM* is parallel to the *x* -axis, we also have

\[\angle FQP = \angle QPM \qquad \quad \dots \left( 2 \right)\]

From (1) and (2), we obtain \(\angle FPQ = \angle QPM\) which means that the tangent at *P* bisects \(\angle FPM.\)

**Example – 19**

Refer to Fig - 26 of the previous example. Prove that angle *PFS* is a right angle.

**Solution:** The equation of the tangent at \(P(a{t^2},2at)\) is

\[ty = x + a{t^2}\]

This intersects the directrix at a point given by \(\left( { - a,\,\,\begin{align}\frac{{a({t^2} - 1)}}{t}\end{align}} \right)\)

The slope of *PF* is

\[{m_{PF}} = \frac{{2at}}{{a{t^2} - a}} = \frac{{2t}}{{{t^2} - 1}}\]

The slope of *SF* is

\[{m_{SF}} = \frac{\begin{align}{\frac{{a({t^2} - 1)}}{t}}\end{align}}{{ - 2a}} = \frac{{a({t^2} - 1)}}{{ - 2t}}\]

Since \({m_{PF}} \times {m_{SF}} = - 1,\) angle *PFS* is a right angle.

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