Question Number 84067 by niroj last updated on 09/Mar/20
$$\:\mathrm{If}\:\boldsymbol{{lx}}+\boldsymbol{{my}}=\mathrm{1}\:\mathrm{touches}\:\mathrm{the}\:\mathrm{curve}\:\left(\boldsymbol{\mathrm{ax}}\right)^{\boldsymbol{\mathrm{n}}} +\left(\boldsymbol{\mathrm{by}}\right)^{\boldsymbol{\mathrm{n}}} =\mathrm{1},\:\mathrm{show}\:\mathrm{that} \\ $$$$\:\left(\frac{\boldsymbol{{l}}}{\boldsymbol{{a}}}\right)^{\frac{\boldsymbol{{n}}}{\boldsymbol{{n}}−\mathrm{1}}} +\left(\frac{\boldsymbol{{m}}}{\boldsymbol{{b}}}\right)^{\frac{\boldsymbol{{n}}}{\boldsymbol{{n}}−\mathrm{1}}} =\mathrm{1}. \\ $$
Answered by mind is power last updated on 09/Mar/20
$${l}={m}={a}={b}\Rightarrow \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\Rightarrow{xy}=\mathrm{0} \\ $$$${the}\:{Quation}\:{is}\:{Tangente}\:? \\ $$$$ \\ $$
Answered by som(math1967) last updated on 09/Mar/20
![let at (p_, q) line touches the curve ∴(ap)^n +(bq)^n =1 slope of line ((dy/dx))_(p,q) =((−a^n p^(n−1) )/(b^n q^(n−1) )) equn. of line y−q=((−a^n p^(n−1) )/(b^n q^(n−1) ))(x−p) a^n p^(n−1) x+b^n q^(n−1) y=(ap)^n +(bq)^n =1 [(ap)^n +(bq)^n =1] Now lx+my=1 and a^n p^(n−1) x+b^n q^(n−1) y=1 are same st. line ∴((a^n p^(n−1) )/l)=((b^n q^(n−1) )/m)=(1/1) ∴p=((l/a^n ))^(1/(n−1)) q=((m/b^n ))^(1/(n−1)) now lp+mq=1 [line touches at (p,q)] ((l×l^(1/(n−1)) )/a^(n/(n−1)) ) +((m×m^(1/(n−1)) )/b^(n/(n−1)) )=1 ((l/a))^(n/(n−1)) +((m/b))^(n/(n−1)) =1](https://www.tinkutara.com/question/Q84095.png)
$${let}\:{at}\:\left({p}_{,} {q}\right)\:{line}\:{touches}\:{the}\:{curve} \\ $$$$\therefore\left({ap}\right)^{{n}} +\left({bq}\right)^{{n}} =\mathrm{1} \\ $$$${slope}\:{of}\:{line} \\ $$$$\:\left(\frac{{dy}}{{dx}}\right)_{{p},{q}} =\frac{−{a}^{{n}} {p}^{{n}−\mathrm{1}} }{{b}^{{n}} {q}^{{n}−\mathrm{1}} } \\ $$$${equn}.\:{of}\:{line} \\ $$$${y}−{q}=\frac{−{a}^{{n}} {p}^{{n}−\mathrm{1}} }{{b}^{{n}} {q}^{{n}−\mathrm{1}} }\left({x}−{p}\right) \\ $$$${a}^{{n}} {p}^{{n}−\mathrm{1}} {x}+{b}^{{n}} {q}^{{n}−\mathrm{1}} {y}=\left({ap}\right)^{{n}} +\left({bq}\right)^{{n}} =\mathrm{1} \\ $$$$\left[\left({ap}\right)^{{n}} +\left({bq}\right)^{{n}} =\mathrm{1}\right] \\ $$$${Now}\:{lx}+{my}=\mathrm{1}\:{and} \\ $$$${a}^{{n}} {p}^{{n}−\mathrm{1}} {x}+{b}^{{n}} {q}^{{n}−\mathrm{1}} {y}=\mathrm{1}\:{are}\:{same} \\ $$$${st}.\:{line}\:\therefore\frac{{a}^{{n}} {p}^{{n}−\mathrm{1}} }{{l}}=\frac{{b}^{{n}} {q}^{{n}−\mathrm{1}} }{{m}}=\frac{\mathrm{1}}{\mathrm{1}} \\ $$$$\therefore{p}=\left(\frac{{l}}{{a}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \:{q}=\left(\frac{{m}}{{b}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \\ $$$${now}\:{lp}+{mq}=\mathrm{1} \\ $$$$\left[{line}\:{touches}\:{at}\:\left({p},{q}\right)\right] \\ $$$$\frac{{l}×{l}^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} }{{a}^{\frac{{n}}{{n}−\mathrm{1}}} }\:+\frac{{m}×{m}^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} }{{b}^{\frac{{n}}{{n}−\mathrm{1}}} }=\mathrm{1} \\ $$$$\left(\frac{{l}}{{a}}\right)^{\frac{{n}}{{n}−\mathrm{1}}} +\left(\frac{{m}}{{b}}\right)^{\frac{{n}}{{n}−\mathrm{1}}} =\mathrm{1} \\ $$
Commented by niroj last updated on 09/Mar/20
$${good}\:{work}. \\ $$
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](https://www.tinkutara.com/question/Q84103.png)
$${my}=\mathrm{1}−{lx} \\ $$$${y}=\left(\frac{−{l}}{{m}}\right){x}+\frac{\mathrm{1}}{{m}} \\ $$$${slope}=\left(\frac{−{l}}{{m}}\right)\:{and}\:{point}\:{of}\:{tangent}\left(\alpha,\beta\right) \\ $$$${l}\alpha+{m}\beta=\mathrm{1}… \\ $$$$\left({ax}\right)^{{n}} +\left({by}\right)^{{n}} =\mathrm{1} \\ $$$${a}^{{n}} ×{nx}^{{n}−\mathrm{1}} +{b}^{{n}} ×{ny}^{{n}−\mathrm{1}} ×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{−{ax}^{{n}−\mathrm{1}} }{{by}^{{n}−\mathrm{1}} }=\left(−\mathrm{1}\right)\left(\frac{{a}}{{b}}\right)\left(\frac{{x}}{{y}}\right)^{{n}−\mathrm{1}} \\ $$$$\left(−\mathrm{1}\right)\left(\frac{{a}}{{b}}\right)\left(\frac{\alpha}{\beta}\right)^{{n}−\mathrm{1}} =\left(\frac{−{l}}{{m}}\right) \\ $$$$\left(\frac{\alpha}{\beta}\right)^{{n}−\mathrm{1}} =\left(\frac{{lb}}{{ma}}\right) \\ $$$$\frac{\alpha}{\beta}=\left(\frac{{lb}}{{ma}}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \\ $$$$\frac{\alpha}{\left({lb}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} }=\frac{\beta}{\left({ma}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} }={k}\:\:\left({say}\right) \\ $$$$\alpha={k}.\left({lb}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \:\:\:\:{and}\:\beta={k}.\left({ma}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \\ $$$${so}\: \\ $$$$\left({a}\alpha\right)^{{n}} +\left({b}\beta\right)^{{n}} =\mathrm{1} \\ $$$$\left\{{a}.{k}.\left({lb}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \right\}^{{n}} +\left\{{b}.{k}.\left({ma}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} \right\}^{{n}} =\mathrm{1} \\ $$$${k}^{{n}} ×\left[{a}^{{n}} .\left({lb}\right)^{\frac{{n}}{{n}−\mathrm{1}}} +{b}^{{n}} .\left({ma}\right)^{\frac{{n}}{{n}−\mathrm{1}}} \right]=\mathrm{1} \\ $$$${l}\alpha+{m}\beta=\mathrm{1} \\ $$$${l}×{k}\left({lb}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} +{m}×{k}\left({ma}\right)^{\frac{\mathrm{1}}{{n}−\mathrm{1}}} =\mathrm{1} \\ $$$$\boldsymbol{{wait}}\:\boldsymbol{{pls}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$