# Examples on Tangents to Parabolas Set 2

**Example – 20**

Refer to Fig - 26 of Example 18 again. Show that angle *FRP* is a right angle.

**Solution:** Once again, we use the equation of the tangent at \(P(a{t^2},2at)\) :

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

This intersects the *y-*axis at the point *R*(*0, at*). The slope of *PR* is simply the slope of the tangent at *P*, i.e. \({m_{PR}} = \frac{1}{t}\)

The slope of *RF* is

\[{m_{RF}} = \frac{{at - 0}}{{0 - a}} = - t\]

Since \({m_{PR}} \times {m_{RF}} = - 1,\) angle *FRP* is a right angle.

The result of the last three examples are important, so we summarize them here :

***** (A) The tangent at any point on a parabola bisects the angle between the focal chord through that point and the perpendicular on the directrix from that point (Example-18).

***** (B) The portion of the tangent to a parabola cut-off between the directrix and the curve subtends a right angle at the focus (Example -19).

***** (C) The perpendicular dropped from the focus onto any tangent to a parabola is concurrent with that tangent and the tangent at the vertex (Example -20)

Along with these, we include a fourth important result here.

***** (D) The tangents at the extremities of any focal chord of a parabola intersect at right angles on the directrix. (Example -16)

**Example – 21**

Three tangents to a parabola form the triangle *PQR*. Prove that the circumcircle of *PQR* passes through the focus of the parabola.

**Solution:** We assume the parabola to be \({y^2} = 4ax\) and three points on it to be \(A(at_1^2,2a{t_1}),\,B(at_2^2,2a{t_2})\) and \(C(at_3^2,2a{t_3}).\) Tangents at *A*, *C* and *B* meet at *P*, *Q* and *R* as shown:

The co-ordinates of the points *P, Q *and *R* are respectively \[(a{t_1}{t_3},\,a({t_1} + {t_2})),\,\,(a{t_2}{t_3},\,a({t_2} + {t_3})\;\rm{}and\;(a{t_1}{t_2},\,a({t_1} + {t_2})).\]

To show that the circumcircle passes through *F*, it would suffice to prove that the chord *PQ* subtends the same angle on *F* as it does on *R*. Since a chord of a circle subtends equal angles anywhere on the circumference, this will prove that *F* also lies on the circumference of the circle.

To evaluate \(\angle PRQ,\) we need the slopes of *PR* and *RQ* which are simply the slopes of the tangents at *A* and *B* respectively, i.e.

\[{m_{PR}} = \frac{1}{{{t_1}}}{\rm{ }} \;and\;\;{\rm{ }}{m_{RQ}} = \frac{1}{{{t_2}}}\]

Thus,

\[\begin{align}\tan (\angle PRQ)&= \frac{{{m_{PR}} - {m_{RQ}}}}{{1 + {m_{PR}}{m_{RQ}}}}\\\\&= \frac{{{t_2} - {t_1}}}{{1 + {t_1}{t_2}}} \qquad \qquad \dots \left( 1 \right)\end{align}\]

Now, to evaluate the angle that the chord *PQ* subtends at *F*, we need to slopes \({m_{PF}}{\rm{ }}\;{\rm{}and}\;{\rm{ }}{m_{QF}}\):

\[\begin{align}&\qquad \quad \;\;{m_{PF}} = \frac{{a({t_1} + {t_3})}}{{a{t_1}{t_3} - a}}\;{\rm{ }}and\;{\rm{ }}{m_{QF}} = \frac{{a({t_2} + {t_3})}}{{a{t_2}{t_3} - a}}\\\\&\Rightarrow \qquad {m_{PF}} = \frac{{{t_1} + {t_3}}}{{{t_1}{t_3} - 1}}{\text{ }}\;and\;{\text{ }}{m_{QF}} = \frac{{{t_2} + {t_3}}}{{{t_2}{t_3} - 1}}\end{align}\]

Thus,

\[\begin{align}& \tan (\angle PFQ) = \frac{{{m_{PF}} - {m_{QF}}}}{{1 + {m_{PF}}{m_{QF}}}}\\\\&\qquad\qquad\quad\;= \frac{{({t_1} + {t_3})({t_2}{t_3} - 1) - ({t_2} + {t_3})({t_1}{t_3} - 1)}}{{({t_1}{t_3} - 1)({t_2}{t_3} - 1) + ({t_1} + {t_3})({t_2} + {t_3})}}\\\\&\qquad\qquad\quad\;= \frac{{{t_2} - {t_1}}}{{1 + {t_1}{t_2}}} \qquad \qquad \qquad \qquad\qquad\qquad\qquad \dots \left( 2 \right)\end{align}\]

Comparing (1) and (2) gives \(\angle PRQ = \angle PFQ\) which confirms that *F* does indeed lie on the circumference of \(\Delta PQR's\) circumcircle.

We could alternatively have done this question by explicitly evaluating the equation of the circle passing through *P*, *Q* and *R* and showing the *F* satisfies that equation.

**Example - 22**

Find the equation of the common tangent(s) to

**(a) ** \({y^2} = 4ax\;\;\text{and}\;\;{(x + a)^2} + {y^2} = {a^2}\)

**(b) **\({y^2} = 4x\;\text{and}\;\;{x^2} = 4y\)

**Solution: (a)** The equation of any tangent to \({y^2} = 4ax\) can be written as

\(ty = x + a{t^2}\)

If this is to touch the circle too, the distance of the circle’s centre, i.e. (a, 0) from this line must be equal to its radius *a*. Thus,

\[\begin{align}& \qquad \frac{{\left| { - a + a{t^2}} \right|}}{{\sqrt {1 + {t^2}} }} = a\\\\& \qquad {(1 - {t^2})^2} = (1 + {t^2})\\\\&\Rightarrow \qquad 1 + {t^4} - 2{t^2} = 1 + {t^2}\\\\&\Rightarrow \qquad {t^2}({t^2} - 3) = 0\\\\&\Rightarrow \qquad t = \pm \sqrt 3 \left( {t = {\rm{ 0\;is\;not\;a\;feasible\;solution\;for\;this\;case}}} \right)\end{align}\]

Thus, the required common tangents are

\[ \pm \sqrt 3 y = x + 3a\]

**(b) ** Any tangent to \({y^2} = 4x\) can be assumed to be of the form

\[ty = x + {t^2}\left( {{\rm{since\; }}{\rm{ }}a = {\rm{ }}1{\rm{\;for\;this\;case}}} \right)\]

If this line is to be tangent to \({x^2} = 4y\) too, its intersection with this curve must yield only one point. Thus, \[t\left( {\frac{{{x^2}}}{4}} \right) = x + {t^2}\] must yield only one root, i.e. its *D* must equal *O*.

This gives

\[\begin{align}& \qquad \;\;16 + 16{t^3} = 0\\&\Rightarrow \quad t = - 1\end{align}\]

The required common tangent is thus

\[x + y + 1 = 0\]

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