रेखा lx + my + n = 0 दीर्घवृत्त \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) के लंब कब होती है?

Question

Question

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

वर्णन:

P(a cos ϕ, b sin ϕ) पर दिए गए दीर्घवृत्त के लंब का समीकरण निम्न है

X a sec ϕ - y b cosec ϕ + b² - a² = 0

इसलिए, lx + my + n = 0 लंब होगा यदि

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

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

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

cos² ϕ + sin²ϕ = 1

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

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