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.


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

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

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


\[px + qy = - r\]

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 :

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