Question

prove that the line x/a+ y/b =1 is tangent to the curve y=be^-x/a at the point, where the curve cuts y-axis.

Answer 1

HELLO DEAR,

the equation of curve is y = be^{-x/a}.
it crosses the y - axis at the point where x = 0.
putting x = 0 in the equation of the curve, we get y = b.
So, the point of the contact is (0 , b).

now, y = be^{-x/a}

dy/dx = -br^{-x/a}/a.

therefore, (dy/dx)(0 , b) = -b/a.

So, the equation of the tangent is

(y - b)/(x - 0) = -b/a

bx + ay = ab

x/a + y/b = 1

Hence, x/a + y/b = 1 touches the curve y = be^{-x/a} at (0,b}

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

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$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \bullet\bf{given \: that}$

$\bf{the \: curve \: \: \: y = {be}^{ \frac{ - x}{a} } }$

$\: \\ \to \bf{it \: meets \: y \: axis \: y = {be}^{0} = b}$

$\: \green{ \bf{ \underline{differentiating \: the \: equation \: wrt \: x}}}$

$\: \qquad \implies \displaystyle \sf{ \frac{dy}{dx} = \bigg({ \frac{ - be}{a} } \bigg)^{ \frac{ - x}{a} } }$

$\: \: \\ \qquad \implies \displaystyle \sf{ \frac{dy}{dx} at(0andb) = { \frac{ - be}{a} }^{0} = - \frac{b}{a} }$

$\: \\ \qquad \implies \underline{\displaystyle \sf{ \: equation \: of \: the \: tangent \:at(b.0) }}$

$\: \qquad \implies \displaystyle \sf{(y - b) = \frac{ - b}{a} (x - 0)}$

$\: \: \qquad \implies \displaystyle \sf{ay - ab = - bx}$

$\: \: \qquad \implies \displaystyle \sf{or \: \: \: \frac{x}{a} + \frac{y}{b} = 1}$

$\green{\underline{\bf{\to hence \: the \: line \: touches \: the \: curve \: at \: the points \: where \: the \: curve \: intersect \: the \: axis \:of y}}}$