First slide

Ellipse

Question

The sum of the squares of the
perpendiculars on any tangent to the ellipse
$x^{2} / a^{2} + y^{2} / b^{2} = 1$ from two points on the minor
axis each at a distance $\sqrt{a^{2} - b^{2}}$ from the centre is

Moderate

A

$2 a^{2}$
B

$2 b^{2}$
C

$a^{2} + b^{2}$
D

$a^{2} - b^{2}$

Solution

The eccentricity e of the given ellipse is given
by $g^{2} = 1 - b^{2} / a^{2} \Rightarrow a^{2} - b^{2} = a^{2} e^{2}$ . So the point on the minor
axis, i.e. y-axis at a distance $\sqrt{a^{2} - b^{2}}$ from the centre $(0, 0)$
of the ellipse are (0, ± ae).
The equation of the tangent at any point $(a \cos θ, b \sin θ)$
on the ellipse is $\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
So the required sum is
$\begin{array}{l} {[\frac{\frac{a e \sin θ}{b} - 1}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}}}]}^{2} + {[\frac{\frac{- a e \sin θ}{b} - 1}{\sqrt{\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}}}]}^{2} \\ = \frac{(a e \sin θ - b)^{2} + (a e \sin θ + b)^{2}}{(b^{2} \cos^{2} θ + a^{2} \sin^{2} θ)} \times a^{2} \\ = \frac{2 a^{2} (a^{2} e^{2} \sin^{2} θ + b^{2})}{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ} \end{array}$
$= \frac{2 a^{2} [(a^{2} - b^{2}) \sin^{2} θ + b^{2})}{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = 2 a^{2}$

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