Example – 9

Let AB be a fixed chord passing through the focus of a parabola. Prove that three circles can be drawn which touch the parabola and AB at the focus.

Solution: Let the parabola be \({y^2} = 4ax;\) its focus is then F(a, 0). The following diagram shows one such circle which touches AB at F and also the parabola :

Let AB have a (fixed) slope m. The equation of AB can be written using the point-slope form :

\[\begin{align}& \qquad \;\;y- 0 = m(x - a)\\\\&\Rightarrow \quad y = mx - am\end{align}\]

Anycircle touching this line at F(a, 0) can be written in terms of a real variable \(\lambda \) as :

\[\begin{align}& \qquad \;\;{(x- a)^2} + {y^2} + \lambda (y - mx + am) = 0\\\\&\Rightarrow \quad{x^2} + {y^2}- (2a + m\lambda )x + \lambda y + a(a + m\lambda ) = 0 \qquad \quad \dots\left( 1 \right)\end{align}\]

We want those values of \(\lambda\) forwhich this circle touches the parabola.

Assume the point of contact of the circle and the parabola to be \(P(a{t^2},2at).\) Thus, at P, the circle and the parabola should have a common tangent.

The tangent at P to the parabola is

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

Sincethis line is a tangent to the circle as well (at the same point \(P(a{t^2},\,2at)),\) we can write the equation of thesame circle using another real parameter \(\alpha :\)

\[\begin{align}& \qquad \;\;{(x- a{t^2})^2} + {(y - 2at)^2} + \alpha (ty - x - a{t^2}) = 0\\\\&\Rightarrow \quad {x^2} + {y^2} - (\alpha + 2a{t^2})x + (\alpha t - 2at)y + {a^2}{t^4} +4{a^2}{t^2} - a\alpha {t^2} = 0 \qquad \quad \dots \left( 2 \right)\end{align}\]

The circles given by (1) and (2) being the same, we can compare the coefficients toobtain the following equations :

\[\begin{align}&2a+ m\lambda = \alpha + 2a{t^2}\\\\&\qquad \lambda = \alpha t - 2at\\\\&a + m\lambda =a{t^4} + 4a{t^2} - \alpha {t^2}\end{align}\]

We can rearrange these equations to make them look more ‘systematic’.

\[\begin{array}{l}(m)\lambda + ( - 1)\alpha + (2a - 2a{t^2}) = 0\\\\(1)\lambda + ( - t)\alpha + (2at) = 0\\\\(m)\lambda + ({t^2})\alpha + (a - 4a{t^2} - a{t^4}) = 0\end{array}\]

The variables \(\lambda \;{\rm{ and}}\;{\rm{ }}\alpha \) and can now be eliminated to obtain arelation purely in terms of t:

\[\left|{\begin{array}{*{20}{c}}m&{ - 1}&{2a(1 - {t^2})}\\1&{ -t}&{2at}\\m&{{t^2}}&{ - a({t^4} + 4{t^2} - 1)}\end{array}} \right|= 0\]

Expanding along \({C_3}\) and simplifying, we obtain

\[\begin{align}& \qquad \;\;mt({t^4}- 2{t^2} - 3) + 1 - 2{t^2} - 3{t^4} = 0\\\\&\Rightarrow \quad mt({t^2} +1)({t^2} - 3) = (3{t^2} - 1)({t^2} + 1)\\\\&\Rightarrow \quad mt({t^2} - 3) =3{t^2} - 1\\\\&\Rightarrow \quad m{t^3} - 3{t^2} - 3mt + 1 = 0 & \dots{\rm{ }}\left(3 \right)\end{align}\]

This is a cubic in t which will have in general three roots. This will imply that three possible points of contact \((a{t^2},\,2at),\) and therefore three possible circles exist satisfying the given property.

But something is missing ! We still have to prove that the cubic will actually yield three real values of t. For that, we follow the approach described in Example - 4

Let \(f(t) = m{t^3} - 3{t^2} - 3mt + 1\)

First,we show that \(f'(t)\) has two real roots, say \({t_1}\;{\rm{and}}\;{t_2}\).

\[\begin{array}{l}f'(t)= 3m{t^2} - 6t - 3m\\\\f'(t) = 0 \Rightarrow m{t^2} - 2t - m = 0\\\\\Rightarrow D = 4 + 4{m^2} > 0\end{array}\]

Thus, \(f'(t)\) has two real and distinct root \({t_1}\;{\rm{and }}\;{t_2}\).

\[ \Rightarrow {t_1} + {t_2} = \frac{2}{m},\,\,\,\,\,{t_1}{t_2} = - 1 \qquad \qquad \dots(4)\]


\[f({t_1})f({t_2})= (mt_1^3 - 3t_1^2 - 3m{t_1} + 1)(mt_2^3 - 3t_2^2 - 3m{t_2} + 1)\] which upon simplification yields (using (4))

\[f({t_1})f({t_2}) = - 4{m^2} - \frac{4}{{{m^2}}} -2\]

which is evidently always negative.

Thus,the cubic (3) will always give three real values of t, and hence three corresponding circles.

You may rest assured that you’ll not encounter a question with this much involved analysis in any exam!This was included here for illustration only.

Example – 10

Let\({C_1}\; {\rm{ and\;}}{C_2}\) be the parabolas \({x^2} = y - 1{\rm{\;\; }}and{\rm{ \;\;}}{y^2} = x - 1\) and respectively. Let P be any point on \({C_1}\) and Q be any point on \({C_2}\). Let \({P_1}\) and \({Q_1}\) bethe reflections of P and Q respectively in the line y = x. Prove that \(PQ \ge \min \{ P{P_1},\,\,Q{Q_1}\} .\) Also, let the points \({P_0}\) and \({Q_0}\) on \({C_1\;\;}and{\rm{ \;\;}}{C_2}\) respectively be such that \({P_0}{Q_0} \le PQ\) for all pairs of points (P, Q) with P on \({C_1}\)and Q on \({C_2}\). Find \({P_0}\)\({Q_0}\).

Solution: The symmetrical nature of the situation tells us that \({P_1}\) will lie on \({C_2}\) and \({Q_1}\) will lie on \({C_1}\):

Observe from the figure that

\[\begin{align}& PO \ge PX\\\\{\rm{and}} \qquad&QO \ge QY\end{align}\]

so that \(PO + QO = PQ \ge PX + QY\)

\[\begin{array}{l}\Rightarrow & 2PQ \ge 2PX + 2QY\\\\\Rightarrow & 2PQ \ge P{P_1} +Q{Q_1}\end{array}\]

Since PQ is greater than the mean of \({PP_1}\)and \({QQ_1}\), it must be greater than (or equal) to the lesser of the two.

To find the minimum possible value of PQ,i.e. \({P_0}\)\({Q_0}\), observe that this length will be minimum when PQ becomes precisely perpendicular to y = x and PQ is normal to both the parabolas at the points of contact P and Q. In such a configuration, assume the co-ordinates of P to be \((t,\,\,{t^2} +1)\) so that Q (which is P's mirror image in such a configuration) will be \(({t^2} + 1,\,t).\) The slope of PQ is this case is –1.Also, the tangent to \({C_1}\) at P will haveslope \(\begin{align}{\left. {\frac{{dy}}{{dx}}}\right|_p} = 2t.\end{align}\) Thus,

\[\begin{align}{}2t\times - 1 = - 1\\\\\Rightarrow t = \frac{1}{2}\end{align}\]

P and Q therefore have the co-ordinate \(\begin{align}\left( {\frac{1}{2},\,\frac{5}{4}} \right) \;{\rm{and}} \;\left({\frac{5}{4},\,\frac{1}{2}} \right)\end{align}\) respectively. The minimum length of PQ is now simply obtained using the distance formula :

\[{P_o}{Q_o} =P{Q_{\min }} = \sqrt {{{\left( {\frac{1}{2} - \frac{5}{4}} \right)}^2} +{{\left( {\frac{5}{4} - \frac{1}{2}} \right)}^2}} = \frac{3}{{2\sqrt 2 }}\]

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