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

Question

A curve $C$ is given by the equation $$e^{\frac{x^2}{2}}y=\ln x^2,$$ where $x\ne0$. Show that $x\ln x^2+e^{\frac{x^2}{2}}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{x}$.[2] 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 $Q$, giving your answer correct to 5 significant figures.[4] Determine the acute angle between the tangents to $C$ at the points $P$ and $Q$.[3]

Accepted Answer

(i) $x\ln x^2+e^{x^2/2}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac2x$ (shown) (ii) $y=\frac2{\sqrt e}(x-1)$, $Q=(-0.39899,-1.6971)$ (iii) $50.2^\circ$ (3 s.f.)

Implicit Curve Tangents and Angle of Intersection

Implicit Curve Tangents and Angle of Intersection

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