Example - 5

Let ABC be an equilateral triangle inscribed in the circle \({x^2} + {y^2} = {a^2}.\) Perpendiculars from A, B, C to the major axis of the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) (where a > b) meet the ellipse respectively at P, Q, R so that P, Q and R lie on the same side of the major axis of A, B and C respectively.Prove that the normals to the ellipse at P, Q and R are concurrent.

Solution: The following figure graphically portrays the situation described in the question:

All the information given about ABC being equilateral and all can be reduced to this single piece of significant information: the polar angles of A, B and C, and hence, the eccentric angles of P, Q and R, will be evenly spaced at \(\frac{{2\pi }}{3},\) by virtue of \(\Delta ABC\) being equilateral.

Thus, we can assume the eccentric angels of P, Q and R to be \(\theta ,\,\,\theta + \frac{{2\pi }}{3},\,\,\theta - \frac{{2\pi }}{3}.\)

Now, the equation of a normal to \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) at eccentric angle \(\phi \) is given by

\[\begin{align}&\qquad ax\sec \phi - by\cos ec\phi = {a^2} - {b^2}.\\& \Rightarrow \quad~~ ax\sin \phi - by\cos \phi = \frac{{({a^2} - {b^2})\sin 2\phi }}{2}.\end{align}\]

Thus, the normals at P, Q and R are respectively given by

\[\begin{align}&{N_P}\quad  : \quad ax\sin \theta - by\cos \theta = \frac{{({a^2} - {b^2})\sin 2\theta }}{2}\\&{N_Q}\quad  :\quad  ax\sin \left( {\theta + \frac{{2\pi }}{3}} \right) - by\cos \left( {\theta + \frac{{2\pi }}{3}} \right) = \frac{{\left( {{a^2} - {b^2}} \right)}}{2}\sin \left( {2\theta + \frac{{4\pi }}{3}} \right)\\&{N_R} \quad :  \quad ax\sin \left( {\theta - \frac{{2\pi }}{3}} \right) - by\cos \left( {\theta - \frac{{2\pi }}{3}} \right) = \frac{{({a^2} - {b^2})}}{2}\sin \left( {2\theta - \frac{{4\pi }}{3}} \right)\end{align}\]

Let us evaluate \(\Delta ,\) the determinant of the coefficients of these three equations:

\[\Delta =\frac{ab({{a}^{2}}-{{b}^{2}})}{2}\left| \begin{align}&\qquad  \sin \theta  &&\qquad \cos \theta  &&\qquad \sin 2\theta   \\&  \sin \left( \theta +\frac{2\pi }{3} \right) && \cos \left( \theta +\frac{2\pi }{3} \right) && \sin \left( 2\theta +\frac{4\pi }{3} \right)  \\& \sin \left( \theta -\frac{2\pi }{3} \right) && \cos \left( \theta -\frac{2\pi }{3} \right) & &\sin \left( 2\theta -\frac{4\pi }{3} \right)  \\\end{align} \right|\]

Using the row operation \({R_1} \to {R_1} + {R_2} + {R_3},\) the first row reduces to zero, which means that

\[\Delta = 0\]

Thus, the normals at P, Q and R must be concurrent.

Example – 6

An ellipse slides between two lines at right angles to one another. Show that the locus of its centre is a circle.

Solution: It would be easiest to assume the two lines to be the coordinate axis, and an ellipse of fixed dimensions sliding between these two lines as shown below:

From the view-point of the ellipse, since the tangents to it at P and Q intersect at right angles at O, the point O must lie on the director circle of the ellipse. Since the radius of the director circle is \(\sqrt {{a^2} + {b^2}} ,\) we must have

\[\begin{align}&\qquad\quad OS = \sqrt {{a^2} + {b^2}} \\ &\Rightarrow \quad O{S^2} = {a^2} + {b^2}\\& \Rightarrow \quad {h^2} + {k^2} = {a^2} + {b^2}\end{align}\]

This must be the locus of the centre S! It can be written in x – y form as

\[{x^2} + {y^2} = {a^2} + {b^2}\]

It is evident that this is a circle centred at the origin and of radius \(\sqrt {{a^2} + {b^2}} .\)

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