Find the value of `n in N` such that the curve `(x/a)^n+(y/b)^n=2` touches the straight line `x/a+y/b=2` at the point `(a , b)dot`

Options

A. `(b,a)`
B. `(a,b)`
C. `(1,1)`
D. ` ((1)/(b),(1)/(a)) `

Answer

B

Explanation

We have,
`(x^(n))/(a^(n))+(y^(n))/(b^(n)) =2 " "...(i) `
Differentiating with respect to x, we get
` n(x^(n-1))/(a^(n)) +n(y^(n-1))/(b^(n))(dy)/(dx)=0 rArr (dy)/(dx)=-(b^(n))/(a^(n))((x)/(y))^(n-1) `
Suppose the line ` (x)/(a)+(y)/(b)=1 ` touches the curve (i) at `P (alpha ,beta).`
The equation of the tangent at `P (alpha, beta )` is
` y- beta =-(b^(n))/(a^(n))((alpha)/(beta))^(n-1) (x-alpha)`
` rArr a^(n) beta^(n-1)y-a^(n)beta^(n)=-b^(n)alpha^(n-1) x+b^(n)alpha^(n)`
` rArr (a^(n)alpha^(n-1))x+(a^(n)beta^(n-1))y=a^(n)beta^(n)+b^(n)alpha^(n) `
` rArr (a^(n)alpha^(n-1))x+(a^(n)beta^(n-1))y=2a^(n)b^(n) " "[because (alpha,beta) " lies on (i)"] `
Comparing this equation with ` (x)/(a)+(y)/(b)=2 ` , we get
` (ab^(n)alpha^(n-1))/(a^(n)b^(n)) =(ba^(n)beta^(n-1))/(a^(n)b^(n))=(2)/(2) `
` rArr ((alpha)/(a))^(n-1)=((beta)/(b))^(n-1)=1 `
` rArr (alpha)/(a)=(beta)/(b)=1 rArr alpha=a,beta =b. `