Question

The locus of the poles of normal chords of an ellipse is given by

• A. $\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}} = \left( a^{2} - b^{2} \right)^{2}$
• B. $\frac{a^{3}}{x^{2}} + \frac{b^{3}}{y^{2}} = \left( a^{2} - b^{2} \right)^{2}$
• C. $\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}} = \left( a^{2} + b^{2} \right)^{2}$
• D. $\frac{a^{3}}{x^{2}} + \frac{b^{3}}{y^{2}} = \left( a^{2} + b^{2} \right)^{2}$

Answer 1

Answer: (A)

$\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}} = \left( a^{2} - b^{2} \right)^{2}$

Answer 2

Let the equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 .....(i)

Let (h, k) be the poles.

Now polar of (h,k) w.r.t the ellipse is given by $\sqrt{1 + \frac{48}{16}} = 2$ = 1

If it is a normal to the ellipse then it must be identical with ax sec θ - by cosec θ = a² – b² ...(iii)

Hence comparing (ii) and (iii), we get

$\frac{\left( h/a^{2} \right)}{a\sec\theta} = \frac{\left( k/b^{2} \right)}{- b\cos ec\theta} = \frac{1}{\left( a^{2} - b^{2} \right)}$⇒ cos θ = $\frac{a^{3}}{h\left( a^{2} - b^{2} \right)}$ and sin θ = $\frac{b^{3}}{k\left( a^{2} - b^{2} \right)}$

Squaring and adding we get, 1 = $\frac{1}{\left( a^{2} - b^{2} \right)^{2}}\left( \frac{a^{6}}{h^{2}} + \frac{b^{6}}{k^{2}} \right)$∴ Required locus of (h, k) is $\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}}$= (a² – b²)².