Question

The equation to the tangent to ${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^n=2$ at $\left(a,b\right)$

• A. $x+y=a+b$
• B. $\frac{x}{a}+\frac{y}{b}=2$
• C. $\frac{x}{a}+\frac{y}{b}=1$
• D. $\frac{x}{a}+\frac{y}{b}=4$

Answer 1

The correct option is A $x+y=a+b$

${\left(\frac{x}{a}\right)}^n+{\left(\frac{y}{b}\right)}^{n}n{\left(\frac{x}{a}\right)}^{n-1}+n{\left(\frac{y}{b}\right)}^{n-1}\frac{dy}{dx}=0$

At point $(a,b)$

$x=a,y=b$

$n+n\frac{dy}{dx}=0\frac{dy}{dx}=-1=my=mx+cy=-x+c$

at$(a,b)$

$b=-a+c$

$c=b+a$

$y+x=b+a$