Q.Prove that the product of the lengths of the perpendiculars drawn from the points $\left(\sqrt{a^2 - b^2},\, 0\right)$ and $\left(-\sqrt{a^2 - b^2},\, 0\right)$ to the line $\dfrac{x}{a}\cos\theta + \dfrac{y}{b}\sin\theta = 1$ is $b^2$.

The key idea is to use the perpendicular distance formula from a point to a line, then multiply the two distances and simplify using the constraint that the points lie on the major axis of an ellipse.

The line can be rewritten as $\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1$, or in standard form:

The perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

For the point $\left(\sqrt{a^2 - b^2}, 0\right)$:

For the point $\left(-\sqrt{a^2 - b^2}, 0\right)$:

The product of the perpendiculars is:

$p_1 \cdot p_2 = \frac{\left|b\sqrt{a^2-b^2}\cos\theta - ab\right| \cdot \left|b\sqrt{a^2-b^2}\cos\theta + ab\right|}{b^2\cos^2\theta + a^2\sin^2\theta}$ …

Why this works: the geometry behind the algebra

The points $(\pm\sqrt{a^2 - b^2}, 0)$ sit symmetrically on the $x$-axis at distance $\sqrt{a^2 - b^2}$ from the origin. If you recognize $a > b$, these are precisely the foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The given line $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$ is a tangent to this ellipse (in parametric form, the tangent at the point $(a\cos\theta, b\sin\theta)$ has exactly this equation).

The beautiful result we're proving is that the product of distances from the two foci to any tangent of an ellipse is constant. This is a deep property of ellipses, but we'll prove it purely algebraically using the distance formula.

Step-by-step proof

1. Set up the distance formula

The perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is:

First, rewrite our line in standard form:

Multiply through by $ab$ to clear denominators:

So $A = b\cos\theta$, $B = a\sin\theta$, $C = -ab$.

2. Calculate the distance from the first focus

For the point $P_1 = (\sqrt{a^2 - b^2}, 0)$:

Factor out $b$ from the numerator:

3. Calculate the distance from the second focus

For the point $P_2 = (-\sqrt{a^2 - b^2}, 0)$:

Factor out $b$:

4. Compute the product …