Animated Solution for Mathematics - Conic Sections: Comprehension Passage

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are witnessing a beautiful dance between two fundamental conic sections: the ellipse and the parabola.

When you look at an equation like

do not just see numbers. See a closed, elegant loop, a planetary orbit, a shape of perfect balance.

And when you see $y^2=4x$, see the open, reaching arms of a parabola, a trajectory of infinite potential. When these two meet, they create a geometry that is rich with symmetry and hidden properties.

Our journey begins by finding where these two worlds collide. We have an ellipse, $\frac{x^2}{9}+\frac{y^2}{8}=1$, and a parabola, $y^2=4x$. The intersection points $M$ and $N$ are the keys to the kingdom.

By substituting the parabola's $y^2$ into the ellipse, we transform a two-variable problem into a single-variable quadratic equation:

This simplifies to $\frac{x^2}{9}+\frac{x}{2}=1$. Multiplying by $18$ clears the path, giving us:

Factoring this is a moment of pure satisfaction: $(2x-3)(x+6)=0$. We discard $x=-6$ because our parabola $y^2=4x$ only exists for $x\ge 0$.

Thus, we are left with $x=\frac{3}{2}$. This gives us our points of intersection: $M(\frac{3}{2},\sqrt{6})$ and $N(\frac{3}{2},-\sqrt{6})$.

Notice the symmetry? The x-coordinates are identical, meaning the chord $MN$ is a vertical line. This symmetry is the heartbeat of this entire problem.

Now, we turn our attention to the triangle $F_{1}MN$. We know $F_1$ is at $(-1,0)$. The orthocenter is the intersection of the altitudes.

In many problems, finding the orthocenter is a tedious algebraic slog. But here, geometry saves us. Because the triangle is symmetric about the x-axis, the altitude from $F_1$ to the base $MN$ is simply the x-axis itself ($y=0$).

To find the orthocenter, we only need one more altitude. Let's take the altitude from $M$ to the side $F_{1}N$. The slope of $F_{1}N$ is:

The altitude must be perpendicular to this, so its slope is the negative reciprocal: $\frac{5}{2\sqrt{6}}$. Using the point-slope form at $M(\frac{3}{2},\sqrt{6})$, we write the equation:

Since the orthocenter lies on the x-axis ($y=0$), we substitute $y=0$ and solve for $x$. The result, $x=-\frac{9}{10}$, is the x-coordinate of our orthocenter. The orthocenter is $(-\frac{9}{10},0)$.

Next, we explore the tangents and normals. The tangent to the ellipse at $M$ is found using the standard formula $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$.

Substituting $M(\frac{3}{2},\sqrt{6})$, we get $\frac{x}{6}+\frac{y\sqrt{6}}{8}=1$. By symmetry, the tangent at $N$ is $\frac{x}{6}-\frac{y\sqrt{6}}{8}=1$. Where do they meet? Set $y=0$, and we find $x=6$. So, $R$ is at $(6,0)$.

Now, the normal to the parabola. The parabola is $y^2=4x$. Differentiating, we get $2y\frac{dy}{dx}=4$, so $\frac{dy}{dx}=\frac{2}{y}$.

At $M$, the slope of the tangent is $\frac{2}{\sqrt{6}}$. The normal is perpendicular, so its slope is $-\frac{\sqrt{6}}{2}$. The equation of the normal at $M$ is:

Setting $y=0$ to find the x-intercept $Q$, we get $2=x-\frac{3}{2}$, which leads to $x=\frac{7}{2}$. Thus, $Q$ is at $(\frac{7}{2},0)$.

We have arrived at the final act: the area ratio. We need the area of triangle $MQR$ and the quadrilateral $M{F}_{1}N{F}_2$.

For $△MQR$, the base lies on the x-axis between $Q(\frac{7}{2},0)$ and $R(6,0)$. The length of the base is $6-\frac{7}{2}=\frac{5}{2}$. The height is the y-coordinate of $M$, which is $\sqrt{6}$.

The area is:

For the quadrilateral $M{F}_{1}N{F}_2$, we use the diagonal method. The diagonals are $F_1{F}_2$ (length $2$) and $MN$ (length $2\sqrt{6}$). Since they are perpendicular, the area is:

Finally, the ratio:

There it is. The ratio is 5:8. You have not just solved a problem; you have mastered a piece of mathematical architecture.