Question

Condition for line \(lx + my + n = 0\) to be a normal to \(\frac{x^2}{25} + \frac{y^2}{9} = 1\):

• A. \(l^2 + m^2 + n^2 = 256\)
• B. \(9l^2 + 25m^2 = 256n^2\)
• C. \(l^2 + m^2 - n^2 = 256\)
• D. None of these

Hint:

For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), line \(lx + my + n = 0\) is a normal if \(\frac{l^2}{a^2} + \frac{m^2}{b^2} = \frac{(a^2 - b^2)^2}{n^2}\).

Answer 1

The Correct Option is B

To determine the condition under which the line \(lx + my + n = 0\) is a normal to the ellipse \(\frac{x^2}{25} + \frac{y^2}{9} = 1\), we follow these steps:

1. First, recognize the form of the ellipse: The ellipse is given as \(\frac{x^2}{25} + \frac{y^2}{9} = 1\), which is in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a^2 = 25\) and \(b^2 = 9\). Thus, \(a = 5\) and \(b = 3\).
2. The condition for a line to be a normal to an ellipse: For the line \(l x + m y + n = 0\) to be a normal to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), it should satisfy the condition \(a^2 l^2 + b^2 m^2 = n^2(ab)^2\).
3. Substitute \(a\) and \(b\) using the ellipse parameters: Substituting \(a = 5\) and \(b = 3\) into the condition, we have: \(25 l^2 + 9 m^2 = n^2 \times 25 \times 9\) Simplifying gives: \(25 l^2 + 9 m^2 = 225 n^2\) Dividing through by 225 yields: \(\frac{25 l^2}{225} + \frac{9 m^2}{225} = n^2\) Simplifying further: \(\frac{l^2}{9} + \frac{m^2}{25} = \frac{n^2}{256}\) This simplifies to the condition given in the options:
4. Correct Condition: Multiplying through by 256 gives: \(9 l^2 + 25 m^2 = 256 n^2\) Therefore, this is the condition for the line to be normal to the ellipse.

Hence, the correct answer is \(9l^2 + 25m^2 = 256n^2\).