Normal at Parametric Point θ

Normal at (a secθ, b tanθ). Slope of normal = −a sinθ/b.

Substitute $x_1 = a\sec\theta$, $y_1 = b\tan\theta$ into the normal $a^2x/x_1 + b^2y/y_1 = a^2+b^2$:

$$\frac{a^2x}{a\sec\theta} + \frac{b^2y}{b\tan\theta} = a^2+b^2$$ $$ax\cos\theta + by\cot\theta = a^2+b^2$$

Multiplying through by $\sin\theta$:

$$ax\sin\theta\cos\theta/\cos\theta \cdot \cos\theta + by\cos\theta = (a^2+b^2)\sin\theta$$

More cleanly, from $ax\cos\theta + by\cot\theta = a^2+b^2$, multiply by $\sin\theta$:

$$ax\sin\theta + by\cos\theta = (a^2+b^2)\sin\theta$$

Hmm — let me redo directly. Normal slope at $(a\sec\theta, b\tan\theta)$ is $-a^2y_1/(b^2x_1) = -a^2b\tan\theta/(b^2a\sec\theta) = -(a\tan\theta)/(b\sec\theta) = -(a\sin\theta)/b$.

Normal through $(a\sec\theta, b\tan\theta)$ with slope $-(a\sin\theta)/b$:

$$y - b\tan\theta = -\frac{a\sin\theta}{b}(x - a\sec\theta)$$ $$by - b^2\tan\theta = -a\sin\theta \cdot x + a^2\sin\theta\sec\theta$$ $$by - b^2\tan\theta = -ax\sin\theta + a^2\tan\theta$$ $$ax\sin\theta + by = a^2\tan\theta + b^2\tan\theta = (a^2+b^2)\tan\theta$$

Therefore:

$$ax\sin\theta + by = (a^2+b^2)\tan\theta$$