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.