42 The inclination angle of the line $l$ is $\frac{\pi}{4}$, and its intercept on the $y$-axis is 3. Using the foci of the hyperbola $12 x^{2}-4 y^{2}=3$ as the foci of an ellipse, construct an ellipse such that the ellipse and $l$ have a common point. To make the major axis of the ellipse have the minimum value, the equation of the ellipse is $\qquad$ $\therefore$.
$42 \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$. The foci of the hyperbola are $F_{1}(1,0), F_{2}(-1,0)$. Suppose the equation of the required ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-1}=1(a \neq 0),
$$

Combining it with $y=x+3$, and eliminating $y$ yields
$$
\left(2 a^{2}-1\right) x^{2}+6 a^{2} x+a^{2}\left(10-a^{2}\right)=0 .
$$

Since $a>c=1$, we have $2 a^{2}-1>0$, and the necessary and sufficient condition for the line to intersect the ellipse is
$$
\left(6 a^{2}\right)^{2}-4\left(2 a^{2}-1\right) a^{2}\left(10-a^{2}\right) \geqslant 0,
$$

Solving this, we get $a^{2} \geqslant 5$. When $a=\sqrt{5}$, the equation of the ellipse is $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$.