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

Answer 1

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.