If lx+my=1 touches the curve (ax)^n +(by)^n =1, show that ((l/a))^(n/(n−1)) +((m/b))^(n/(n−1)) =1.


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Question Number 84067 by niroj last updated on 09/Mar/20

$If l x + m y = 1 touches the curve {(ax)}^{n} + {(by)}^{n} = 1, show that$ ${(\frac{l}{a})}^{\frac{n}{n - 1}} + {(\frac{m}{b})}^{\frac{n}{n - 1}} = 1 .$
	Answered by mind is power last updated on 09/Mar/20

	$l = m = a = b \Rightarrow$ $1 + 1 = 2$ $x + y = 1$ $x^{2} + y^{2} = 1 \Rightarrow x y = 0$ $t h e Q u a t i o n i s T a n g e n t e ?$
	Answered by som(math1967) last updated on 09/Mar/20

	$l e t a t (p_{,} q) l i n e t o u c h e s t h e c u r v e$ $∴ {(a p)}^{n} + {(b q)}^{n} = 1$ $s l o p e o f l i n e$ ${(\frac{d y}{d x})}_{p, q} = \frac{- a^{n} p^{n - 1}}{b^{n} q^{n - 1}}$ $e q u n . o f l i n e$ $y - q = \frac{- a^{n} p^{n - 1}}{b^{n} q^{n - 1}} (x - p)$ $a^{n} p^{n - 1} x + b^{n} q^{n - 1} y = {(a p)}^{n} + {(b q)}^{n} = 1$ $[{(a p)}^{n} + {(b q)}^{n} = 1]$ $N o w l x + m y = 1 a n d$ $a^{n} p^{n - 1} x + b^{n} q^{n - 1} y = 1 a r e s a m e$ $s t . l i n e ∴ \frac{a^{n} p^{n - 1}}{l} = \frac{b^{n} q^{n - 1}}{m} = \frac{1}{1}$ $∴ p = {(\frac{l}{a^{n}})}^{\frac{1}{n - 1}} q = {(\frac{m}{b^{n}})}^{\frac{1}{n - 1}}$ $n o w l p + m q = 1$ $[l i n e t o u c h e s a t (p, q)]$ $\frac{l \times l^{\frac{1}{n - 1}}}{a^{\frac{n}{n - 1}}} + \frac{m \times m^{\frac{1}{n - 1}}}{b^{\frac{n}{n - 1}}} = 1$ ${(\frac{l}{a})}^{\frac{n}{n - 1}} + {(\frac{m}{b})}^{\frac{n}{n - 1}} = 1$
	Commented by niroj last updated on 09/Mar/20

	$g o o d w o r k .$
	Answered by TANMAY PANACEA last updated on 09/Mar/20
	$my=1−lx y=(((−l)/m))x+(1/m) slope=(((−l)/m)) and point of tangent(α,β) lα+mβ=1... (ax)^n +(by)^n =1 a^n ×nx^(n−1) +b^n ×ny^(n−1) ×(dy/dx)=0 (dy/dx)=((−ax^(n−1) )/(by^(n−1) ))=(−1)((a/b))((x/y))^(n−1) (−1)((a/b))((α/β))^(n−1) =(((−l)/m)) ((α/β))^(n−1) =(((lb)/(ma))) (α/β)=(((lb)/(ma)))^(1/(n−1)) (α/((lb)^(1/(n−1)) ))=(β/((ma)^(1/(n−1)) ))=k (say) α=k.(lb)^(1/(n−1)) and β=k.(ma)^(1/(n−1)) so (aα)^n +(bβ)^n =1 {a.k.(lb)^(1/(n−1)) }^n +{b.k.(ma)^(1/(n−1)) }^n =1 k^n ×[a^n .(lb)^(n/(n−1)) +b^n .(ma)^(n/(n−1)) ]=1 lα+mβ=1 l×k(lb)^(1/(n−1)) +m×k(ma)^(1/(n−1)) =1 wait pls$
	$m y = 1 - l x$ $y = (\frac{- l}{m}) x + \frac{1}{m}$ $s l o p e = (\frac{- l}{m}) a n d p o i n t o f t a n g e n t (α, β)$ $l α + m β = 1 . . .$ ${(a x)}^{n} + {(b y)}^{n} = 1$ $a^{n} \times {n x}^{n - 1} + b^{n} \times {n y}^{n - 1} \times \frac{d y}{d x} = 0$ $\frac{d y}{d x} = \frac{- {a x}^{n - 1}}{{b y}^{n - 1}} = (- 1) (\frac{a}{b}) {(\frac{x}{y})}^{n - 1}$ $(- 1) (\frac{a}{b}) {(\frac{α}{β})}^{n - 1} = (\frac{- l}{m})$ ${(\frac{α}{β})}^{n - 1} = (\frac{l b}{m a})$ $\frac{α}{β} = {(\frac{l b}{m a})}^{\frac{1}{n - 1}}$ $\frac{α}{{(l b)}^{\frac{1}{n - 1}}} = \frac{β}{{(m a)}^{\frac{1}{n - 1}}} = k (s a y)$ $α = k . {(l b)}^{\frac{1}{n - 1}} a n d β = k . {(m a)}^{\frac{1}{n - 1}}$ $s o$ ${(a α)}^{n} + {(b β)}^{n} = 1$ ${a . k . {(l b)}^{\frac{1}{n - 1}}}^{n} + {b . k . {(m a)}^{\frac{1}{n - 1}}}^{n} = 1$ $k^{n} \times [a^{n} . {(l b)}^{\frac{n}{n - 1}} + b^{n} . {(m a)}^{\frac{n}{n - 1}}] = 1$ $l α + m β = 1$ $l \times k {(l b)}^{\frac{1}{n - 1}} + m \times k {(m a)}^{\frac{1}{n - 1}} = 1$ $w a i t p l s$
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