Two tangents to an exponential-logarithmic curve · source reference

Printed question

A curve $C$ is given by the equation $e^{x^2}y=\ln x^2$, where $x\ne0$.

(i) Show that $2x\ln x^2+e^{x^2}\frac{dy}{dx}=\frac2x$. [2]

(ii) Find the equation of the tangent to $C$ at the point $P$ where $x=1$. Given that the tangent at $P$ meets $C$ again at the point $Q$, find the coordinates of $Q$, leaving your answer correct to 5 significant figures. [4]

(iii) Determine the acute angle, in degrees, between the tangents to $C$ at the points $P$ and $Q$. [3]

[2022/HCI/Prelim/I/9]

Printed working

(i) Differentiate both sides of $e^{x^2}y=\ln x^2$ with respect to $x$:

$$2xe^{x^2}y+e^{x^2}\frac{dy}{dx}=\frac1{x^2}(2x)\Rightarrow2x\ln x^2+e^{x^2}\frac{dy}{dx}=\frac2x$$

(shown), because $e^{x^2}y=\ln x^2$.

(ii) When $x=1$, $y=\ln1^2/e^{1^2}=0$; $\frac{dy}{dx}=\frac{2/1-2(1)\ln1^2}{e^{1^2}}=2/e$. Tangent at $P$: $y=\frac2e(x-1)$.

At the next intersection, $\frac2e(x-1)=\frac{\ln x^2}{e^{x^2}}$. Using GC, the other intersection is $Q(-0.49516,-1.1001)$ (5 s.f.).

(iii) At $Q$, $\left.\frac{d}{dx}[\ln(x^2)/e^{x^2}]\right|_{x=-0.49516}=-4.250281$ (GC); alternatively substitute $x=-0.49516$ into (i).

$\alpha=\tan^{-1}(2/e)=36.344^\circ$, $\beta=\tan^{-1}(4.250281)=76.760^\circ$.

Required acute angle $=180^\circ-36.344^\circ-76.760^\circ=66.9^\circ$ (1 d.p.).

Teacher notes

Apply product rule on $e^{x^2}y$: $\frac d{dx}(e^{x^2}y)=y\frac d{dx}(e^{x^2})+e^{x^2}\frac{dy}{dx}$.

Let $\alpha$ be the acute angle that the tangent at $P$ makes with the positive $x$-axis. $\tan\alpha=\text{gradient of tangent at }P=(dy/dx)_P=2/e$.

Let $\beta$ be the acute angle that the tangent at $Q$ makes with the positive $x$-axis. $\tan\beta=|\text{gradient at }Q|=|(dy/dx)_Q|=4.250281$.

Answer key

(ii) $y=\frac2e(x-1)$; $Q(-0.49516,-1.1001)$ (5 s.f.). (iii) $66.9^\circ$ (1 d.p.).

Native diagrams