Question

The equation of the normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$at the positive end of the latus-rectum is

• A. $x + ey + e^{3}a = 0$
• B. $x - ey - e^{3}a = 0$
• C. $x - ey - e^{2}a = 0$
• D. None of these

Answer 1

Answer: (B)

$x - ey - e^{3}a = 0$

Answer 2

The equation of the normal at $(x_{1},y_{1})$ to the given ellipse is $\frac{a^{2}x}{x_{1}} - \frac{b^{2}y}{y_{1}} = a^{2} - b^{2}$. Here, $x_{1} = ae$ and $y_{1} = \frac{b^{2}}{a}$

So, the equation of the normal at positive end of the latus- rectum is

$\frac{a^{2}x}{ae} - \frac{b^{2}y}{b^{2}/a} = a^{2}e^{2}$ [$\because b^{2} = a^{2}(1 - e^{2})$] ⇒ $\frac{ax}{e} - ay = a^{2}e^{2}$

⇒$x - ey - e^{3}a = 0$