Show that the product of the lengths of its perpendicular segments drawn from the foci to any tangent line to the ellipse x^2/25 + y^2/16 = 1

Question

Show that the product of the lengths of its perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac {x^2}{25} + \frac {y^2}{16} = 1\) is equal to 16.

x²/25 + y²/16 = 1

Answer 1

Given equation of the ellipse is  \(\frac {x^2}{25} + \frac {y^2}{16} = 1\)

Comparing this equation with  \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) we get

a² = 25 and b² = 16

∴ a² = 25, b² = 16

∴ a = 5, b = 4

We know that e = \(\frac {\sqrt{a^2-b^2}}{a}\)

∴ e =\(\frac {\sqrt{25-16}}{5}\)= = 3/5

ae = 5 (3/5) = 3

Co-ordinates of foci are S(ae, 0) and S'(-ae, 0), i.e., S(3, 0) and S'(-3, 0) 

Equations of tangents to the ellipse \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) having slope m are

y = mx ± \(\sqrt{a^2 m^2 + b^2}\)

Equation of one of the tangents to the ellipse is

y = mx + \(\sqrt{25 m^2 + 16}\)

∴ mx – y +\(\sqrt{25 m^2 + 16}\)= 0…..(i)

p₁ = length of perpendicular segment from S(3, 0) to the tangent (i)

p₂ = length of perpendicular segment from S'(-3, 0) to the tangent (i)

= 16