Show that the line x-a-y-b-1 touches the curve y-b - e-x-a at the point where the curve intersects the axis of Y-

We have the equation of line given by x/a + y/b = 1 which touches the curve y =b.e^ x/a at the point where the curve intersects the axis of Y i.e., x = 0 ∴ y ...

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Byju's Answer

Standard XII

Mathematics

Directrix of Hyperbola

Show that the...

Question

Show that the line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve y = $b . e^{- \frac{x}{a}}$ at the point where the curve intersects the axis of Y.

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Solution

We have the equation of line given by $\frac{x}{a} + \frac{y}{b} = 1$ which touches the curve y = $b . e^{- \frac{x}{a}}$ at the point where the curve intersects the axis of Y i.e., x = 0
$∴ y = b . e^{- \frac{x}{a}}$
So, the point of intersection of the curve with Y - axis is (0,b).
Now, slope of the given line at (0,b) is given by
$\frac{1}{a} .1 + \frac{1}{b} . \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{1}{a} . b \Rightarrow \frac{d y}{d x} = - \frac{1}{a} . b = \frac{- b}{a} = m_{1}$
Also, the slope of the curve at (0,b) is
$\frac{d y}{d x} = b . e^{- \frac{x}{a}} . \frac{- 1}{a} \frac{d y}{d x} = \frac{- b}{a} e^{- \frac{x}{a}} {(\frac{d y}{d x})}_{(0, b)} = \frac{- b}{a} e^{- 0} = \frac{- b}{a} = m_{2} S i n c e, m_{1} = m_{2} = \frac{- b}{a}$
Hence the line touches the curve at the point where the curve intersects the axis of y.

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Show that the line xa+yb=1 touches the curve y = b.e−xa at the point where the curve intersects the axis of Y.

Show that the line $\frac{x}{a} + \frac{y}{b} = 1$ touches the curve y = $b . e^{- \frac{x}{a}}$ at the point where the curve intersects the axis of Y.