Question

Let e be the eccentricity of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). If a=5, b=4 and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is \(lx+my=27\), then l+m=

• A. \( \frac{3}{e} \)
• B. \( \frac{3}{2e} \)
• C. \( \frac{6}{e} \)
• D. \( \frac{1}{e} \)

Hint:

Memorize the equation of the normal to an ellipse at \((x_1, y_1)\): \( \frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 \). This is much faster than finding the tangent's slope, then the normal's slope, and using the point-slope form.

Answer 1

The Correct Option is C

Step 1: Find the parameters of the ellipse and the coordinates of the end of the latus rectum.
Given \(a=5\) and \(b=4\). So \(a^2=25\) and \(b^2=16\).
Eccentricity \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}\).
The coordinates of the end of the latus rectum in the first quadrant are \( (ae, b^2/a) \).
The point is \( (5 \cdot \frac{3}{5}, \frac{16}{5}) = (3, \frac{16}{5}) \).
Step 2: Find the equation of the normal at this point.
The equation of the normal to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at \( (x_1, y_1) \) is \( \frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 \).
Substitute the values: \(a^2=25, b^2=16, x_1=3, y_1=16/5\).
\( \frac{25x}{3} - \frac{16y}{16/5} = 25 - 16 \).
\( \frac{25x}{3} - 5y = 9 \).
To match the form \(lx+my=27\), we multiply the equation by 3.
\( 25x - 15y = 27 \).
Step 3: Identify l and m and find their sum.
By comparing \(25x - 15y = 27\) with \(lx+my=27\), we get \(l=25\) and \(m=-15\).
The sum is \(l+m = 25 + (-15) = 10\).
Step 4: Check which option equals 10.
We have \(e=3/5\).
Option (C) is \( \frac{6}{e} = \frac{6}{3/5} = 6 \cdot \frac{5}{3} = 10 \).
This matches our result.