Question

The line lx + my + n = 0 is a normal to the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) if
1. \(\frac{{{a^2}}}{{{m^2}}} + \frac{{{b^2}}}{{{l^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}\)
2. \(\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}\)
3. \(\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}\)
4. None of these

Answer 1

Correct Answer - Option 2 : \(\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}\)

Concept:

Equation of any normal is \(ax - sec θ - by\ cosec θ = a^2 - b^2\)    ...(i)

Calculation:

Equation of line, lx + my = n is normal to the ellipse     ... (ii)

⇒ my = -lx + n

\(⇒ y = -\dfrac{l}{m} x + \dfrac{n}{m}\)

Here Slope m = - l/m and c = n/m

Equation of ellipse,

\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}=1\)

So, Equation (i) and (ii) represents the same line:

⇒ a secθ/b cosecθ × m = (a² - b²)/-n 

⇒ cosθ = -an/l(a2 - b2)

And, sinθ = bn/m(a2 - b2)

∵ cos²θ + sin²θ = 1

∴ [-an/l(a2 - b2)]² + [bn/m(a2 - b2)]² = 1

 \(⇒\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}\)

This is the required condition