Question

Find the equation of tangent to y=be ⁻ˣ/ᵃ where it intersects Y-axis.

Answer 1

we have to find equation of tangent to y=be ⁻ˣ/ᵃ where it intersects Y-axis.

if curve intersects at Y-axis, x = 0
so, y=$be^{0/a}$ = b
hence, intersecting point on Y-axis by curve is (0,b)

now, $y=be^{-x/a}$
differentiate with respect to x,
$\frac{dy}{dx}=-\frac{b}{a}e^{-x/a}$

at (0,b) , $\frac{dy}{dx}|_{(0,b)}$ = -b/a

hence, slope of tangent of curve = $\frac{dy}{dx}|_{(0,b)}$ = -b/a

now, equation of tangent at (0,b) :
(y - b) = -b/a(x - 0)

a(y - b) + bx = 0

ay - ab + bx = 0

bx + ay - ab = 0

therefore, equation of tangent of curve is bx + ay - ab = 0

Answer 2

Dear Student:

y=be ⁻ˣ/ᵃ

Equation of y axis, x=0

Intersection of curve and y axis.

$y=bexp(\frac{-x}{a})=bexp(\frac{-0}{a})=b$

So,point is (0,b)

Now slope of tangent is m

m=dy/dx

$\frac{dy}{dx} =b\frac{d}{dx}exp(-x/a)$

$m=\frac{-b}{a}*exp(\frac{-x}{a})$

at (0,b) slope m is = -b/a

So, equation of tangent passing through the (0,b)

$(y-y_{0} )=m(x-x_{0})$

$(x_{0},y_{0} )=(0,b)$

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

$(y-b )=-b/a *x$