Question

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

Answer 1

Given equation of the ellipse is ${\displaystyle \frac{x^2}{25}+\frac{y^2}{16}}$ = 1.

 Comparing this equation with ${\displaystyle \frac{x^2}{{\text{a}}^2}+\frac{y^2}{{\text{b}}^2}}$ = 1, we get

∴ a² = 25, b² = 16

∴ a = 5, b = 4

We know that e = ${\displaystyle \frac{\sqrt{{\text{a}}^2-{\text{b}}^2}}{\text{a}}}$

∴ e = ${\displaystyle \frac{\sqrt{25-16}}{5}}$

= ${\displaystyle \frac{\sqrt{9}}{5}}$

= ${\displaystyle \frac{3}{5}}$

ae = ${\displaystyle 5\left(\frac{3}{5}\right)}$

= 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

${\displaystyle \frac{x^2}{{\text{a}}^2}+\frac{y^2}{{\text{b}}^2}}$ = 1 having slope m are

y = ${\displaystyle \text{m}x\pm \sqrt{{\text{a}}^2{\text{m}}^2+{\text{b}}^2}}$

Equation of one of the tangents to the ellipse is

y = ${\displaystyle \text{m}x+\sqrt{25{\text{m}}^2+16}}$

∴ ${\displaystyle \text{m}x-y+\sqrt{25{\text{m}}^2+16}}$ = 0   ...(i)

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

= ${\displaystyle \left|\frac{\text{m}\left(3\right)-0+\sqrt{25{\text{m}}^2+16}}{\sqrt{{\text{m}}^2+1}}\right|}$

∴ p₁ = ${\displaystyle \left|\frac{3\text{m}+\sqrt{25{\text{m}}^2+16}}{\sqrt{{\text{m}}^2+1}}\right|}$

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

= ${\displaystyle \left|\frac{\text{m}(-3)-0+\sqrt{25{\text{m}}^2+16}}{\sqrt{{\text{m}}^2+1}}\right|}$

∴ p₂ = ${\displaystyle \left|\frac{-3\text{m}+\sqrt{25{\text{m}}^2+16}}{\sqrt{{\text{m}}^2+1}}\right|}$

∴ p₁p₂ = ${\displaystyle \left|\frac{3\text{m}+\sqrt{25{\text{m}}^2+16}}{\sqrt{{\text{m}}^2+1}}\right|\left|\frac{-3\text{m}+\sqrt{25{\text{m}}^2+16}}{\sqrt{{\text{m}}^2+1}}\right|}$

= ${\displaystyle \frac{(25{\text{m}}^2+16)-9{\text{m}}^2}{{\text{m}}^2+1}}$

= ${\displaystyle \frac{16({\text{m}}^2+1)}{{\text{m}}^2+1}}$

= 16