At which point the line x-a-y-b-1- touches the curve y - be -x - aA- 0-0B- 0- aC- 0- bD- b- 0

The correct option is C (0, b)Let the point be (x_1,y_1), ∴ y_1=be^ x_1/a.....(i) Also, curve y=be^ x/a⇒dy/dx= b/ae^ x/a (dy/dx )_(x_1,y_1)= b/ae^ x_1/a= b/a ...

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Byju's Answer

Standard XII

Mathematics

Definition of Circle

At which poin...

Question

At which point the line $\frac{x}{a} + \frac{y}{b} = 1$ , touches the curve $y = b e^{- x / a}$

(0, 0)

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(0,a)

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(0, b)

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(b, 0)

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Solution

The correct option is C (0, b)
Let the point be $(x_{1}, y_{1}), ∴ y_{1} = b e^{- x_{1} / a} . . . . . (i)$
Also, curve $y = b e^{- x / a} \Rightarrow \frac{d y}{d x} = \frac{- b}{a} e^{- x / a}$
${(\frac{d y}{d x})}_{(x_{1}, y_{1})} = \frac{- b}{a} e^{- x_{1} / a} = \frac{- b}{a}$
$\Rightarrow x_{1} = 0$ .
Hence, the point is (0, b).

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At which point the line xa+yb=1, touches the curve y=be−x/a

The correct option is C (0, b)Let the point be (x1,y1),∴y1=be−x1/a.....(i) Also, curve y=be−x/a⇒dydx=−bae−x/a (dydx)(x1,y1)=−bae−x1/a=−ba ⇒x1=0. Hence, the point is (0, b).

At which point the line $\frac{x}{a} + \frac{y}{b} = 1$ , touches the curve $y = b e^{- x / a}$