The line lx + my + n = 0 is a normal to the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) if

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This question was previously asked in

AAI ATC Junior Executive 2015 Official Paper (Shift 1)

\(\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{{{n^2}}}\)
\(- \frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} - \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{{{n^2}}} = 0\)
\(\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{{{n^2}}}\)
\(\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} + \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{{{n^2}}} = 0\)

Answer 1

Explanation:

The equation of the normal of the given ellipse at P(a cos ϕ, b sin ϕ) is

X a sec ϕ - y b cosec ϕ + b² - a² = 0

So, lx + my + n = 0 will be a normal if

\(\frac{l}{{a\sec \phi }} = \frac{m}{{ - b\;cosec\;\phi }} = \frac{n}{{{b^2} - {a^2}}} \)

\(\Rightarrow \cos \phi = \frac{{na}}{{l\left( {{b^2} - {a^2}} \right)}} \)

\(\sin \phi = \frac{{ - nb}}{{m\left( {{b^2} - {a^2}} \right)}} \)

cos² ϕ + sin²ϕ = 1

\(\Rightarrow \frac{{{n^2}{a^2}}}{{{L^2}{{\left( {{b^2} - {a^2}} \right)}^2}}} + \frac{{{n^2}{b^2}}}{{{m^2}\left( {{b^2} - {a^2}} \right)}} = 1 \)

\(\Rightarrow \frac{{{a^2}}}{{{L^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{{{n^2}}} \)