Show that xp - yq - 1 touches the curve y - qe-x-p at the point- where the curve crosses the axis of y-axis-

[ y=q e^ x / p #160;; ⇒ dy dx = q p e^ x / p #160;; when the curves crosses t ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Equation of Tangent at a Point (x,y) in Terms of f'(x)

Show that x...

Question

Show that $\frac{x}{p} + \frac{y}{q} = 1$ touches the curve $y = q e^{\frac{- x}{p}}$ at the point, where the curve crosses the axis of $y -$ axis.

Open in App

Solution

$\begin{matrix} y = q e^{- \frac{x}{p}} \Rightarrow \frac{d y}{d x} = - \frac{q}{p} e^{- \frac{x}{p}} w h e n t h e c u r v e s c r o s s e s t h e y - a x i s, x = 0 A t x = 0, y = q e^{- \frac{0}{p}}, i . e . y = q S o, t h e c u r v e c r o s s e s t h e y - a x i s a t (0, q) i s y - q = - \frac{q}{p} (x - 0) \frac{x}{p} + \frac{y}{q} = 1 \end{matrix}$
Hence, proved.

Suggest Corrections

Similar questions

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.

Q. Show that the line

\frac{x}{a} + \frac{y}{b} = 1

, tocuhes the curve

y = b . e^{\frac{- x}{a}}

at the point where the curve intersects the axis of

y

Q. The curve

y = a x^{3} + b x^{2} + c x + 5

touches the x -axis at P(-2,0) and cuts the y-axis at the point Q, where its gradient is 3. Find the equation of the curve completely.

Q. The curve

y = a x^{3} + b x + 5

touches the x-axis at

P (- 2, 0)

and cuts y-axis at point

Q

where its gradient is

3.

Find

a

and

b .

Also find coordinates of point

Q .

Q. the curve

y = a x^{3} + b x^{2} + c x + 5

touches the x-axis at P(-2,0) and cuts the y-axis at a point Q where its gradient is 3. Find a, b, c.

Join BYJU'S Learning Program

Show that xp+yq=1 touches the curve y=qe−xp at the point, where the curve crosses the axis of y−axis.

y=qe−xp⇒dydx=−qpe−xpwhenthecurvescrossesthey−axis,x=0Atx=0,y=qe−0p,i.e.y=qSo,thecurvecrossesthey−axisat(0,q)isy−q=−qp(x−0)xp+yq=1Hence, proved.

Show that $\frac{x}{p} + \frac{y}{q} = 1$ touches the curve $y = q e^{\frac{- x}{p}}$ at the point, where the curve crosses the axis of $y -$ axis.