Question

answer this question of advance maths​

Answer 1

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}} </p><p>$

Let

$\frac{l}{x} = \frac{m}{y} = \frac{n}{z} = k \:$

So,

$l \: = kx \: , m \: = ky \: , \: n \: = kz.........(1)$

It is given that

$\displaystyle \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } + \frac{ {z}^{2} }{ {c}^{2} } = 1 ..........(2)$

LHS

$= \displaystyle \: \frac{ {l}^{2} }{ {a}^{2} } + \frac{ {m}^{2} }{ {b}^{2} } + \frac{ {n}^{2} }{ {c}^{2} }$

$= \displaystyle \: \frac{ {k}^{2} {x}^{2} }{ {a}^{2} } + \frac{ {k}^{2} {y}^{2} }{ {b}^{2} } + \frac{ {k}^{2} {z}^{2} }{ {c}^{2} } \: \: \: \: \: using(1)$

$= {k}^{2} ( \displaystyle \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } + \frac{ {z}^{2} }{ {c}^{2} })$

$= {k}^{2} \times 1 \: \: \: \: using(2)$

$= {k}^{2}$

RHS

$\displaystyle \: \frac{ {l}^{2} + {m}^{2} + {n}^{2} }{ {x}^{2} + {y}^{2} + {z}^{2} }$

$= \displaystyle \: \frac{ {k}^{2} {x}^{2} + {k}^{2} {y}^{2} + {k}^{2} {z}^{2} }{ {x}^{2} + {y}^{2} + {z}^{2} } \: \: using(1)$

$= \displaystyle \: \frac{ {k}^{2} ({x}^{2} + {y}^{2} + {z}^{2})}{{x}^{2} + {y}^{2} + {z}^{2}}$

$= {k}^{2}$

Hence

LHS = RHS

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