# Examples On Normals To Ellipses Set-2

**Example – 34**

The normal at any point P on the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) meets the major and minor axes at A and B respectively. ON is the perpendicular upon this normal from the centre O of the ellipse. Show that

\[PA \cdot PN = {b^2}andPB \cdot PN = {a^2}\]

**Solution:**

Assume the point P to be \((a\cos \theta ,\,b\sin \theta ).\) The normal at P has the equation

\[(a\sec \theta )x - (b\,{\rm{cosec}}\,\theta )y = {a^2} - {b^2}\qquad\qquad...\left( 1 \right)\]

The coordinates of A are therefore

\[A \equiv \left( {\frac{{{a^2} - {b^2}}}{{a\sec \theta }},\,0} \right) \equiv (a{e^2}\cos \theta ,\,0)\]

Similarly, B is

\[B \equiv \left( {0,\,\,\frac{{{b^2} - {a^2}}}{{b\,{\rm{cosec }}\theta }}} \right) \equiv \left( {0,\,\,\frac{{ - {a^2}{e^2}}}{b}\sin \theta } \right)\]

PA and PB can now be evaluated using the distance formula:

\[\begin{align}&\;\;PA = \sqrt {{{(a\cos \theta - a{e^2}\cos \theta )}^2} + {{(b\sin \theta )}^2}} \\&\qquad{\rm{ }} = \sqrt {\frac{{{b^4}}}{{{a^2}}}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \\&\qquad{\rm{ }} = \frac{b}{a}\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \qquad\qquad...\left( 2 \right)\\&\;\;PB = \sqrt {{{(a\cos \theta )}^2} + {{\left( {b\sin \theta + \frac{{{a^2}{e^2}}}{b}\sin \theta } \right)}^2}} \\&\qquad{\rm{ }} = \frac{a}{b}\sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }\qquad\qquad...\left( 3 \right)\end{align}\]

PN can be evaluated either using the perpendicular distance of O from the normal at \(P(P{N^2} = O{P^2} - O{N^2})\) or simply as the perpendicular distance of O from the tangent at P.

The tangent at P has the equation

\[bx\cos \theta + ay\sin \theta - ab = 0\]

Thus,

\[PN = \frac{{\left| {ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}\qquad\qquad...\left( 4 \right)\]

From (2), (3) and (4), we have

\[PA \cdot PN = {b^2}\;and\;PB \cdot PN = {a^2}\]

**Example - 35**

Prove that from any given point\(P(h,\,k),\) four normals (real or imaginary) can be drawn to the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) and the sum of the eccentric angles of the feet of these four normals is an odd integral multiple of \(\pi \)

**Solution:** Any normal to the ellipse is of the form

\[(a\sec \theta )x - (b\,{\rm{cosec }}\theta )y = {a^2} - {b^2}\]

If this passes through \(P(h,\,\,k),\) we have

\[ah\sec \theta - bk\,{\rm{cosec }}\theta = {a^2} - {b^2}\qquad...\left( 1 \right)\]

We need to show that this equation will in general yield four values of \(\theta \) For this purpose, we use the substitution

\[\cos \theta \to \frac{{1 - {t^2}}}{{1 + {t^2}}},\,\,\,\sin \theta = \frac{{2t}}{{1 + {t^2}}}\]

where \(t = \tan \frac{\theta }{2}.\) Thus, (1) transforms to,

\[\begin{align}&ah\left( {\frac{{1 + {t^2}}}{{1 - {t^2}}}} \right) - bk\left( {\frac{{1 + {t^2}}}{{2t}}} \right) = {a^2} - {b^2} = {a^2}{e^2}(\because \,\,{b^2} = {a^2}(1 - {e^2})) \hfill \\&\Rightarrow \quad bk{t^4} + 2(ah + {a^2}{e^2}){t^3} + 2(ah - {a^2}{e^2})t - bk = 0\quad\qquad...\left( 2 \right) \hfill \\\end{align} \]

This is a biquadratic equation in t, yielding four roots, say \({t_1},\,{t_2},\,\,{t_3},\,\,{t_4}.\) This shows that in general four normals can be drawn.

From (2), we have

\[\begin{align}&{s_1} = {t_1} + {t_2} + {t_3} + {t_4} = - \frac{{2(ah + {a^2}{e^2})}}{{bk}}\\&{s_2} = {t_1}{t_2} + {t_1}{t_3} + {t_1}{t_4} + {t_2}{t_3} + {t_2}{t_4} + {t_3}{t_4} = 0\\&{s_3} = {t_1}{t_2}{t_3} + {t_1}{t_2}{t_4} + {t_1}{t_3}{t_4} + {t_2}{t_3}{t_4} = - \frac{{2(ah - {a^2}{e^2})}}{{bk}}\\&{s_4} = {t_1}{t_2}{t_3}{t_4} = \frac{{ - bk}}{{bk}} = - 1\end{align}\]

Thus,

\[\begin{align}&\tan \left( {\frac{{{\theta _1}}}{2} + \frac{{{\theta _2}}}{2} + \frac{{{\theta _3}}}{2} + \frac{{{\theta _4}}}{2}} \right) = \frac{{{s_1} - {s_3}}}{{1 - {s_2} + {s_4}}} = \infty \\ &\Rightarrow \quad\frac{{{\theta _1}}}{2} + \frac{{{\theta _2}}}{2} + \frac{{{\theta _3}}}{2} + \frac{{{\theta _4}}}{2} = \left( {n + \frac{1}{2}} \right)\pi \\ & \Rightarrow\quad {\theta _1} + {\theta _2} + {\theta _3} + {\theta _4} = (2n + 1)\pi\end{align}\]

This proves that the sum of the eccentric angles is an odd multiple of \(\pi \)

**TRY YOURSELF – IV**

**Q. 1** The normal at any point P on an ellipse cuts its major axis in Q. Show that the locus of the mid-point of PQ is an ellipse.

**Q. 2 ** If the eccentric angles of points A and B on the ellipse are \(\phi \;and\;\phi + \frac{\pi }{2}\;and\;\theta \) and and is the angle between the normals at A and B, prove that the eccentricity e of the ellipse is given by \(\frac{2}{{{e^2}}}\sqrt {1 - {e^2}} = {\sin ^2}2\phi {\tan ^2}\theta .\)

**Q. 3** The tangent drawn at the point \(({t^2},\,2t)\) on the parabola \({y^2} = 4x\) is the same as the normal drawn at a point \((\sqrt 5 \cos \theta ,\,\,2\sin \theta )\) on the ellipse \(4{x^2} + 5{y^2} = 20.\) What are the values of t and \(\theta \,\,?\)

**Q. 4 ** If the normal at an end of a latus rectum of an ellipse passes through are extremity of the minor axis, show that its eccentricity e satisfies \(1 - {e^2} = {e^4}.\)

**Q. 5 (a)** A ray is incident on the ellipse \(16{x^2} + 9{y^2} = 400\) at a point with y-coordinate 4. The source of this ray is at (–3, 0). Find the equation of the reflected ray.

** (b)** Prove that the normal at any point on an ellipse bisects the angles between the focal radii of that point.

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school