Determine the parameter m so that the curve y² = mx is

Determine the parameter $m$ so that the curve $y^2 = mx$ is tangent to the curve $y = e^x$, and calculate the area bounded by these curves and the $y$-axis.

We will first pin down $m$ from the tangency conditions, then compute the enclosed area.

Part 1: Find the parameter $m$

Step 1: Express both curves as $x$ in terms of $y$

Parabola: $y^2=mx \implies x=\frac{y^2}{m}$ (for $m>0$). Exponential: $y=e^x \implies x=\ln y$.

Step 2: Impose tangency — equal derivatives

\[\frac{dx}{dy}\Big|_{\text{parabola}} = \frac{2y}{m}, \qquad \frac{dx}{dy}\Big|_{\text{exponential}} = \frac{1}{y}\]

At the point of tangency $y=y_0$:

\[\frac{2y_0}{m} = \frac{1}{y_0} \implies m = 2y_0^2\]

Step 3: Impose tangency — equal $x$-values

\[\frac{y_0^2}{m} = \ln y_0\]

Substituting $m=2y_0^2$:

\[\frac{y_0^2}{2y_0^2} = \ln y_0 \implies \frac12 = \ln y_0 \implies y_0 = e^{1/2} = \sqrt{e}\]

Step 4: Solve for $m$

\[m = 2y_0^2 = 2e\]

\[\boxed{m = 2e}\]

Step 5: Verify

At $y_0=\sqrt e$: parabola gives $x=\frac{e}{2e}=\frac12$; exponential gives $x=\ln\sqrt e=\frac12$. ✓ Point of tangency: $\left(\frac12,\sqrt e\right)$.

Part 2: Calculate the area

Step 1: Set up the region

The parabola is $x=\frac{y^2}{2e}$ and the exponential is $x=\ln y$; they touch at $\left(\frac12,\sqrt e\right)$ and the parabola also passes through the origin $(0,0)$. The exponential curve, however, only approaches the $y$-axis as $y\to0^+$ (since $x=\ln y\to-\infty$) and never actually reaches it at a finite point — so the finite region bounded by the $y$-axis, the parabola, and the tangency condition is the area swept out by the parabola alone, from $y=0$ up to the point of tangency $y=\sqrt e$.

Step 2: Set up and evaluate the integral

\[A = \int_0^{\sqrt e} \frac{y^2}{2e}\,dy = \frac{1}{2e}\left[\frac{y^3}{3}\right]_0^{\sqrt e} = \frac{1}{2e}\cdot\frac{e^{3/2}}{3} = \frac{\sqrt e}{6}\]

\[\boxed{A = \frac{\sqrt{e}}{6} \text{ square units}}\]

Step 3: Numerical value

\[A = \frac{\sqrt e}{6} \approx 0.275 \text{ square units}\]

Summary

Parameter:

\[\boxed{m = 2e}\]

Area:

\[\boxed{A = \frac{\sqrt{e}}{6} \approx 0.275 \text{ square units}}\]

Key insight: Internal tangency between two curves expressed as $x=f(y)$ gives two equations (matching position and matching slope) that pin down both the free parameter $m$ and the point of contact simultaneously.