# Ellipse – Normal

## Ellipse – Normal

Definition: Let S = 0 be an ellipse and P be a point on the ellipse S = 0. The line passing through P and perpendicular to tangent of S = 0 at P is called the normal to the ellipse S = 0 at P.

Theorem: The equation of the normal to the ellipse $$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$$ at P (x₁, y₁) is $$\frac{{{a}^{2}}x}{{{x}_{1}}}-\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}-{{b}^{2}}$$ .

Proof: The equation of the tangent to S = 0 at P is S₁ = 0

$$\frac{{{x}_{1}}x}{{{a}^{2}}}+\frac{{{y}_{1}}y}{{{b}^{2}}}-1=0$$.

The equation of the normal to S = 0 at P is $$\frac{{{y}_{1}}}{{{b}^{2}}}(x-{{x}_{1}})-\frac{{{x}_{1}}}{{{a}^{2}}}(y-{{y}_{1}})=0$$.

$$\frac{x{{y}_{1}}}{{{b}^{2}}}-\frac{y{{x}_{1}}}{{{a}^{2}}}=\frac{{{x}_{1}}{{y}_{1}}}{{{b}^{2}}}-\frac{{{x}_{1}}{{y}_{1}}}{{{a}^{2}}}$$.

$$\frac{{{a}^{2}}{{b}^{2}}}{{{x}_{1}}{{y}_{1}}}\left( \frac{x{{y}_{1}}}{{{b}^{2}}}-\frac{y{{x}_{1}}}{{{a}^{2}}} \right)=\frac{{{a}^{2}}{{b}^{2}}}{{{x}_{1}}{{y}_{1}}}\left( \frac{{{x}_{1}}{{y}_{1}}}{{{b}^{2}}}-\frac{{{y}_{1}}{{x}_{1}}}{{{a}^{2}}} \right)$$.

$$\frac{{{a}^{2}}x}{{{x}_{1}}}-\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}-{{b}^{2}}$$.

Example: Find the equation of the normal to the ellipse x² + 8y² = 33 at (-1, 2).

Solution: Given

Method 1: x² + 8y² = 33 at (-1, 2)

The equation of the tangent to the ellipse S = 0 at P (x₁, y₁) is S₁ = 0

xx₁ + 8yy₁ = 33 at (-1, 2)

x₁ = -1, y₁ = 2

x (-1) + 8y (2) = 33

– x + 16y = 33

x – 16y + 33 = 0

Equation of the normal at (-1, 2) is 16x + y + k = 0

16 (-1) + (2) + k = 0

k = 14

the equation of the normal is 16x + y + 14 = 0

16x + y + 14

Method 2: The equation of the normal to the ellipse $$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$$ at P (x₁, y₁) is $$\frac{{{a}^{2}}x}{{{x}_{1}}}-\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}-{{b}^{2}}$$.

x² + 8y² = 33

$$\frac{{{x}^{2}}}{33}+\frac{{{y}^{2}}}{\frac{33}{8}}=1$$ … (1)

$$\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$$ … (2)

From equation 1 and 2

a² = 33 and b² = 33/8

$$\frac{{{a}^{2}}x}{{{x}_{1}}}-\frac{{{b}^{2}}y}{{{y}_{1}}}={{a}^{2}}-{{b}^{2}}$$.

$$\frac{33x}{-1}-\frac{\left( \frac{33}{8} \right)y}{2}=33-\frac{33}{8}$$ .

$$-\frac{66x}{2}-\frac{(33)y}{16}=\frac{231}{8}$$.

$$-\frac{528x}{16}-\frac{(33)y}{16}=\frac{231}{8}$$.

– 528x – 33y = 462

16x + y + 14.