Question

A curve is represented parametrically by the equations

x = t + e^at and y = – t + e^at when

r ∈ R and a > 0. If the curve touches the axis of x at the point A, then the coordinates of the point A are –

• A. (1, 0)
• B. (1/e, 0)
• C. (e, 0)
• D. (2e, 0)

Answer 1

Answer: (D)

(2e, 0)

Answer 2

x = t + e^at; y = – t + e^at

$\frac{dx}{dt}$ = 1 + ae^at ; $\frac{dy}{dt}$ = – 1 + ae^at

$\frac{dy}{dx}$ = $\frac{–1 + ae^{at}}{1 + ae^{at}}$

At the point A, y = 0 and $\frac{dy}{dx}$ = 0 for some t = t₁

∴ $ae^{at_{1}}$ = 1 …(i)

Also, 0 = – t₁ + $e^{at_{1}}$; ∴ $e^{at_{1}}$ = t₁ …(ii)

On putting this value in Eq. (i), we get

at₁ = 1 ⇒ t₁ = $\frac{1}{a}$;

Now, from Eq. (i), ae = 1 ⇒ a = $\frac{1}{e}$

Hence, x_A = t₁ +$e^{at_{1}}$= e + e = 2e

⇒ A ≡ (2e, 0).