# Examples On Normals To Ellipses Set-1

**Example - 32**

What is the farthest distance at which a normal to the ellipse \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) can lie from the centre of the ellipse ?

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

\[ax\sec \theta - by\,{\rm{cosec }}\theta = {a^2} - {b^2}\]

The distance of this normal from the centre (0, 0) is

\[d = \frac{{\left| {{a^2} - {b^2}} \right|}}{{\sqrt {{a^2}{{\sec }^2}\theta + {b^2}\,{\rm{cose}}{{\rm{c}}^2}\,\theta } }}\]

We need to find the maximum value of d, or equivalently, the minimum value of

\[f(\theta ) = {a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\,\theta \]

We have

\[\begin{align} & {f}'(\theta )=2{{a}^{2}}{{\sec }^{2}}\theta \tan \theta -2{{b}^{2}}\text{cose}{{\text{c}}^{2}}\theta \cot \theta \\ & \Rightarrow\quad {f}'(\theta )=0~when\;{{\tan }^{2}}\theta =\frac{b}{a} \\ &\qquad\qquad\qquad\quad \Rightarrow\quad \tan \theta =\pm \sqrt{\frac{b}{a}} \\ \end{align}\]

Verify that at this value of \(\theta ,f''(\theta )\) is positive so that this \(\theta \) indeed gives us the minimum value of \(f(\theta ).\)

Now,

\[\begin{align}&\qquad{f_{\min }}(\theta ) = {a^2}{\sec ^2}{\theta _{\min }} + {b^2}{\rm{cose}}{{\rm{c}}^2}\,{\theta _{\min }}\\&\qquad\quad\qquad{\rm{ }} = {a^2}(1 + {\tan ^2}{\theta _{\min }}) + {b^2}(1 + {\cot ^2}{\theta _{\min }})\\&\qquad\quad\qquad{\rm{ }} = {(a + b)^2}\\& \;\;\Rightarrow \quad {d_{\max }} = \frac{{\left| {{a^2} - {b^2}} \right|}}{{\sqrt {{f_{\min }}(\theta )} }}\\&\qquad\quad\qquad{\rm{ }} = \frac{{\left| {{a^2} - {b^2}} \right|}}{{\left( {a + b} \right)}}\\&\qquad\quad\qquad{\rm{ }} = \left| {a - b} \right|\end{align}\]

**Example – 33**

Find the point on the ellipse \(\frac{{{x^2}}}{6} + \frac{{{y^2}}}{3} = 1\) whose distance from the line \(x + y = 7\) is minimum.

**Solution:** Any point on the given ellipse can be assumed to be \(P(\sqrt 6 \cos \theta ,\,\sqrt 3 \sin \theta ).\) From the following figure, observe that for the distance of P from the given line to be minimum, the normal at P must be perpendicular to the given line.

The equation of the normal at P, using parametric form, is

\[(\sqrt 6 \sec \theta )x - (\sqrt 3 {\rm{cosec}}\,\theta )y = 3\]

whose slope is

\[{m_N} = \sqrt 2 \tan \theta \]

If the normal is perpendicular to \(x + y = 7,\) we have

\[\begin{align}&\quad\qquad{m_N} = 1\\ &\Rightarrow \quad \tan \theta = \frac{1}{{\sqrt 2 }}\\ &\Rightarrow \quad\sin \theta = \frac{1}{{\sqrt 3 }}\;and\;\cos \theta = \frac{{\sqrt 2 }}{{\sqrt 3 }}\end{align}\]

Thus, the point P is \((a\cos \theta ,\,b\sin \theta ) \equiv (2,\,1).\)