Question

The locus of the poles of normal chords of the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 is-

• A. $\frac{a^{6}}{x^{2}} - \frac{b^{6}}{y^{2}}$= (a² + b²)²
• B. $\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}}$ = (a² – b²)²
• C. $\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}}$= (a² + b²)²
• D. None of these

Answer 1

Answer: (B)

$\frac{a^{6}}{x^{2}} + \frac{b^{6}}{y^{2}}$ = (a² – b²)²

Answer 2

Let (a, b) be the pole of the normal

ax sec f – by cosec f = a² – b² … (1)

Then the pole of (a, b) with respect to the ellipse viz, $\frac{x\alpha}{a^{2}} + \frac{y\beta}{b^{2}}$= 1 … (2)

Must represent the same straight line as (1).

On comparing equations (1) and (2), we get

$\frac{a^{3}\sec\varphi}{\alpha}$ = = a² – b²

\ cos f = $\frac{a^{3}}{\alpha(a^{2} - b^{2})}$ and sin f = $\frac{b^{3}}{\beta(a^{2} - b^{2})}$

On squaring and adding , we have

1 = $\frac{a^{6}}{\alpha^{2}(a^{2} - b^{2})^{2}}$ + $\frac{b^{6}}{\beta^{2}(a^{2} - b^{2})^{2}}$

Ž $\frac{a^{6}}{\alpha^{2}}$ + $\frac{b^{6}}{\beta^{2}}$ = (a² – b²)².

Hence the locus of (a, b) is$\frac{a^{6}}{x^{2}}$ + $\frac{b^{6}}{y^{2}}$ = (a² – b²)².

Hence (2) is the correct answer.