Prove that the sum of the squares of the perpendiculars on any tangent from two points on the minor axis, each distant $\sqrt{a^{2} - b^{2}}$ from the centre, is $2 a^{2}$ .

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Solution

The equation of tangent to ellipse is $b x cos θ + a y sin θ = a b$
The two points on minor axis is $(0, \sqrt{a^{2} - b^{2}})$ and $(0, - \sqrt{a^{2} - b^{2}})$
The length of perpendicular from those two points on tangent are
$\frac{a (\sqrt{a^{2} - b^{2}} sin θ - b)}{\sqrt{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}}$ & $\frac{a (- \sqrt{a^{2} - b^{2}} sin θ - b)}{\sqrt{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}}$
By using $∣ ∣ ∣ \frac{A x + B y + C}{\sqrt{A^{2} + B^{2}}} ∣ ∣ ∣$
So, sum of square of perpendiculars is $= {(\frac{a (\sqrt{a^{2} - b^{2}} sin θ - b)}{\sqrt{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}})}^{2} + {(\frac{a (- \sqrt{a^{2} - b^{2}} sin θ - b)}{\sqrt{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}})}^{2}$
$= \frac{a^{2} (\sqrt{a^{2} - b^{2}} sin θ - b)^{2}}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ} + \frac{a^{2} (- \sqrt{a^{2} - b^{2}} sin θ - b)^{2}}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}$
$= \frac{a^{2} ((a^{2} - b^{2}) {sin}^{2} θ + b^{2} - 2 \sqrt{(a^{2} - b^{2})} sin θ . b + (a^{2} - b^{2}) {sin}^{2} θ + b^{2} + 2 \sqrt{a^{2} - b^{2}} sin θ)}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}$
$= \frac{a^{2} (2 (a^{2} - b^{2}) {sin}^{2} θ + 2 b^{2})}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}$
$= \frac{2 a^{2} (a^{2} {sin}^{2} θ - b^{2} {sin}^{2} θ + 2 b^{2})}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}$
$= \frac{2 a^{2} (a^{2} {sin}^{2} θ + b^{2} (1 - {sin}^{2} θ))}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ}$
$= \frac{2 a^{2} (a^{2} {sin}^{2} θ + b^{2} {cos}^{2} θ)}{b^{2} {cos}^{2} θ + a^{2} {sin}^{2} θ} = 2 a^{2}$

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Prove that the sum of the squares of the perpendiculars on any tangent from two points on the minor axis, each distant √a2−b2 from the centre, is 2a2.

Prove that the sum of the squares of the perpendiculars on any tangent from two points on the minor axis, each distant $\sqrt{a^{2} - b^{2}}$ from the centre, is $2 a^{2}$ .