Question:

**Step 1: Determine the equation of the normal.** A line with equal positive intercepts on the x and y-axes has the form $x+y=c$, where $c>0$. **Step 2: Apply the normal condition to the hyperbola.** The given hyperbola is $\frac{x^2}{4} - \frac{y^2}{1} = 1$. Comparing with the standard form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, we have $A^2 = 4$ and $B^2 = 1$. The equation of the normal to the hyperbola at a point $(x_0, y_0)$ is given by: $$ \frac{A^2 x}{x_0} + \frac{B^2 y}{y_0} = A^2 + B^2 $$ Substituting the values of $A^2$ and $B^2$: $$ \frac{4x}{x_0} + \frac{y}{y_0} = 4 + 1 $$ $$ \frac{4x}{x_0} + \frac{y}{y_0} = 5 $$ The slope of this normal is $m_N = -\frac{4/x_0}{1/y_0} = -\frac{4y_0}{x_0}$. The slope of the line $x+y=c$ is $m_L = -1$. For the line $x+y=c$ to be the normal, their slopes must be equal: $$ -\frac{4y_0}{x_0} = -1 $$ $$ 4y_0 = x_0 $$ **Step 3: Use the hyperbola equation to find the point of tangency.** Since the point $(x_0, y_0)$ lies on the hyperbola, it satisfies its equation: $$ \frac{x_0^2}{4} - y_0^2 = 1 $$ Substitute $x_0 = 4y_0$ into the hyperbola equation: $$ \frac{(4y_0)^2}{4} - y_0^2 = 1 $$ $$ \frac{16y_0^2}{4} - y_0^2 = 1 $$ $$ 4y_0^2 - y_0^2 = 1 $$ $$ 3y_0^2 = 1 $$ $$ y_0^2 = \frac{1}{3} \implies y_0 = \pm \frac{1}{\sqrt{3}} $$ Since $x_0 = 4y_0$, we have $x_0 = \pm \frac{4}{\sqrt{3}}$. The normal $x+y=c$ has positive intercepts, which implies $c>0$. The normal passes through $(x_0, y_0)$, so $x_0+y_0=c$. If $y_0 = \frac{1}{\sqrt{3}}$ and $x_0 = \frac{4}{\sqrt{3}}$, then $c = \frac{4}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}$. This value of $c$ is positive. If $y_0 = -\frac{1}{\sqrt{3}}$ and $x_0 = -\frac{4}{\sqrt{3}}$, then $c = -\frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}} = -\frac{5}{\sqrt{3}}$. This value of $c$ is negative, which contradicts the condition of positive intercepts. Therefore, we must have $x_0 = \frac{4}{\sqrt{3}}$, $y_0 = \frac{1}{\sqrt{3}}$, and $c = \frac{5}{\sqrt{3}}$. **Step 4: Apply the tangency condition to the ellipse.** The normal line is $x+y=c$, where $c = \frac{5}{\sqrt{3}}$. This line touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The condition for a line $lx+my=n$ to be tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $n^2 = a^2 l^2 + b^2 m^2$. For the line $x+y=c$, we have $l=1$, $m=1$, and $n=c$. Thus, the tangency condition becomes: $$ c^2 = a^2(1)^2 + b^2(1)^2 $$ $$ c^2 = a^2 + b^2 $$ Substitute the value of $c = \frac{5}{\sqrt{3}}$: $$ \left(\frac{5}{\sqrt{3}}\right)^2 = a^2 + b^2 $$ $$ \frac{25}{3} = a^2 + b^2 $$ **Step 5: Calculate the required expression.** The problem asks for the value of $3(a^2 + b^2)$. $$ 3(a^2 + b^2) = 3 \left(\frac{25}{3}\right) $$ $$ 3(a^2 + b^2) = 25 $$