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Question
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
Solution
Given that y = b · e– x/a, the equation of curve and `x/"a" + y/"b"` = 1, the equation of line.
Let the coordinates of the point where the curve intersects the y-axis be (0, y1)
Now differentiating y = b · e– x/a both sides w.r.t. x, we get
`"dy"/"dx" = "b" * "e"^((-x)/"a") (- 1/"a")`
= `- "b"/"a" * "e"^((-x)/"a")`
So, the slope of the tangent, m1 = `- "b"/"a" * "e"^((-x)/"a")`
Differentiating `x/"a" + y/"b"` = 1 both sides w.r.t. x, we get
`1/"a" + 1/"b" * "dy"/"dx"` = 0
So, the slope of the line, m2 = ` (-"b")/"a"`.
If the line touches the curve, then m1 = m2
⇒ `(-"b")/"a" * "e"^((-x)/"a") = (-"b")/"a"`
⇒ `"e"^((-x)/"a")` = 1
⇒ `(-x)/"a" log "e"` = log 1 .....(Taking log on both sides)
⇒ `(-x)/"a"` = 0
⇒ x = 0
Putting x = 0 in equation y = `"b" * "e"^((-x)/"a")`
⇒ y = b · e0 = b
Hence, the given equation of curve intersects at (0, b) i.e. on y-axis.
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