# Normals To Ellipses

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Consider an ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$$ We need to find the equation of the normal to this ellipse at a given point P on it. In general, we also need to find what condition must be satisfied if $$y = mx + c$$ is to be a normal to this ellipse.

NORMAL AT P(x1, y1) : The slope of the tangent at P can be found by evaluating the derivative $$\frac{{dy}}{{dx}}$$ at P.

Thus,

${m_T} = {\left. {\frac{{dy}}{{dx}}} \right|_P} = {\left. {\frac{{ - {b^2}x}}{{{a^2}y}}} \right|_P} = \frac{{ - {b^2}{x_1}}}{{{a^2}{y_1}}}$

Therefore, the slope of the normal is

${m_N} = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}$

The equation of the normal can now be written using point-slope form:

\begin{align}&y - {y_1} = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}(x - {x_1}) \hfill \\&\Rightarrow \quad \boxed{\frac{{{a^2}}}{{{x_1}}}x - \frac{{{b^2}}}{{{y_1}}}y = {a^2} - {b^2}} \hfill \\\end{align}

NORMAL AT $$P(a\,cos\theta ,\,\,b sin\theta )$$:If P has been specified in parametric form, the equation for the normal can be obtained by the substitution $${x_1} \to a\cos \theta ,\,\,{y_1} \to b\sin \theta$$ in the equation obtained above :

$\boxed{ax\sec \theta - by\,{\text{cosec }}\theta = {a^2} - {b^2}}$

This form of the normal is the most widely used. It represents the normal at a point whose eccentric angle is $$\theta$$

NORMAL OF SLOPE m : Let $$y = mx + c$$ be a normal to the ellipse (say at the point$$P({x_1},\,\,{y_1})).$$

The equation of the normal at $$P({x_1},\,\,{y_1})$$ can be written as

$\frac{{{a^2}x}}{{{x_1}}} - \frac{{{b^2}y}}{{{y_1}}} = {a^2} - {b^2}{\rm{ }}\qquad\qquad...\left( 1 \right)$

This has a slope  \begin{align}m = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}\qquad\qquad...\left( 2 \right)\end{align}

Since $$P({x_1},\,\,{y_1})$$ lies on the ellipse, we have

$\frac{{x_1^{\,\,2}}}{{{a^2}}} + \frac{{y_1^{\,\,2}}}{{{b^2}}} = 1\qquad\qquad...\left( 3 \right)$

Using (2) in (3), we obtain

\begin{align}&\frac{{x_1^{\,\,2}}}{{{a^2}}} + \frac{{{b^2}{m^2}}}{{{a^4}}}x_1^2 = 1\\ &\Rightarrow \quad{x_1} = \pm \frac{{{a^2}}}{{\sqrt {{a^2} + {b^2}{m^2}} }}\qquad\qquad...\left( 4 \right)\end{align}

From (2), ${y_1} = \pm \frac{{{b^2}m}}{{\sqrt {{a^2} + {b^2}{m^2}} }}\qquad\qquad...\left( 5 \right)$

Using (4) and (5) in (1), we finally obtain the equation of the normal with slope m as

$y = mx \pm \frac{{m({a^2} - {b^2})}}{{\sqrt {{a^2} + {b^2}{m^2}} }}\qquad\qquad...\left( 6 \right)$

We can say that any line of the form (6) will be a normal to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,$$ with the points of contact given by (4) and (5).

Example - 31

When is the straight line $$px + qy + r = 0$$ a normal to the ellipse \begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,?\end{align}

Solution: Any normal to the ellipse can be written using the parametric form as

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

If

$px + qy = - r$

is also a normal to the ellipse, we have

\begin{align}&\quad\quad\frac{{a\sec \theta }}{P} = - \frac{{b\,{\rm{cosec }}\theta }}{q} = \frac{{{a^2} - {r^2}}}{{ - r}}\\ &\;\Rightarrow\quad \cos \theta = \frac{{ - ar}}{{p({a^2} - {b^2})}},\,\,\,\sin \theta = \frac{{br}}{{q({a^2} - {b^2})}}\end{align}

Eliminating $$\theta$$ gives us the required condition :

\begin{align}&{\cos ^2}\theta + {\sin ^2}\theta = 1\\ &\Rightarrow \quad \frac{{{{({a^2} - {b^2})}^2}}}{{{r^2}}} = \frac{{{a^2}}}{{{p^2}}} + \frac{{{b^2}}}{{{q^2}}}\end{align}