If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas

Q: Why did we get $x=0$ and $x=4a$? Is there a shortcut to finding these intersection points?

Great observation! Since the parabolas are symmetric about the line $y=x$, you could have set $x^2 = 4ax$ (by replacing $y$ with $x$ in the first equation), which gives $x^2 - 4ax = 0$, leading directly to $x=0$ and $x=4a$. Both methods are valid, but the substitution method is more robust for general cases.

Q: Why does $d$ have to be zero?

Because the origin $(0,0)$ lies on the line $2bx + 3cy + 4d = 0$. If a line passes through the origin, its constant term must be zero. Substituting $(0,0)$ into the equation gives $2b(0) + 3c(0) + 4d = 0$, which simplifies to $4d = 0$, hence $d=0$.

Q: Why do we square the equations at the end? Is that necessary?

We square them to combine the two conditions ($d=0$ and $2b+3c=0$) into a single, elegant equation that equals zero, which matches the format of the given options. It's a common technique in coordinate geometry to consolidate multiple constraints into one.

Question

If $a 
eq 0$ and the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of the parabolas $y^2 = 4ax$ and $x^2 = 4ay$, then

Options:

(A) $d^2 + (3b - 2c)^2 = 0$

(B) $d^2 + (3b + 2c)^2 = 0$

(C) $d^2 + (2b - 3c)^2 = 0$

(D) $d^2 + (2b + 3c)^2 = 0$

Sigma Percentile · Accepted Answer

### The Dance of the Parabolas: A Geometric Journey

Welcome, future engineer! Today, we are going to explore a problem that might look like a dry algebraic exercise, but is actually a beautiful dance between two geometric curves.

We are looking at the intersection of two parabolas, $y^2 = 4ax$ and $x^2 = 4ay$. Imagine these two curves on your coordinate plane. One is a classic parabola opening to the right, and the other is its mirror image, opening upwards.

They meet at the origin, and they meet at another point in the first quadrant. Our mission is to find the relationship between the coefficients of a line that passes through these two points of intersection.

### Phase 1: The Algebraic Hunt

To find where these curves meet, we must solve their equations simultaneously. We have $y^2 = 4ax$ and $x^2 = 4ay$.

Let's take the second equation and express $y$ in terms of $x$: 
$$y = \frac{x^2}{4a}$$

Now, we substitute this into the first equation. This is where the algebra gets interesting:
$$\left( \frac{x^2}{4a} ight)^2 = 4ax$$

Expanding this, we find:
$$\frac{x^4}{16a^2} = 4ax$$

Multiplying both sides by $16a^2$, we arrive at $x^4 = 64a^3x$. Rearranging this, we get $x^4 - 64a^3x = 0$.

Factoring out $x$, we have $x(x^3 - 64a^3) = 0$. This gives us two real roots: $x = 0$ and $x = 4a$. These are the $x$-coordinates of our intersection points!

### Phase 2: Finding the Points

Now that we have our $x$-coordinates, finding the $y$-coordinates is a breeze. When $x = 0$, $y = 0$. So, our first point is the origin $(0, 0)$.

When $x = 4a$, substituting back into $y = \frac{x^2}{4a}$ gives:
$$y = \frac{(4a)^2}{4a} = 4a$$

So, our second intersection point is $(4a, 4a)$. We have successfully mapped the geometry of the intersection.

### Phase 3: The Line Constraint

We are told that the line $2bx + 3cy + 4d = 0$ passes through these two points. This means both points must satisfy the line's equation.

Let's start with the origin $(0, 0)$. Substituting $x=0$ and $y=0$ into the line equation, we get $2b(0) + 3c(0) + 4d = 0$, which simplifies to $4d = 0$.

Since $4$ is not zero, $d$ must be $0$. This is a massive simplification!

Now, with $d=0$, our line equation becomes $2bx + 3cy = 0$. We substitute our second point $(4a, 4a)$ into this equation:
$$2b(4a) + 3c(4a) = 0$$

Factoring out $4a$, we get $4a(2b + 3c) = 0$. Since the problem guarantees $a 
eq 0$, we can safely divide by $4a$. This leaves us with the elegant relation $2b + 3c = 0$.

### Phase 4: The Final Synthesis

We have two results: $d = 0$ and $2b + 3c = 0$. To match the options provided in the problem, we square both equations to get $d^2 = 0$ and $(2b + 3c)^2 = 0$.

Adding them together, we get:
$$d^2 + (2b + 3c)^2 = 0$$

This is the final, beautiful result. It perfectly matches the fourth option.

You see? By breaking the problem down into the geometry of the curves and the algebra of the line, we turned a complex-looking problem into a simple, logical path. Keep practicing this mindset, and you will master coordinate geometry!

Animated Solution for Mathematics - Conic Sections: If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then

Select Answer:

Visualized Solution

Visualizing the Parabolas & Line

Setting up the Intersection Equations

Substituting and Expanding

Solving the Equation for $x$

Finding the Intersection Coordinates

Substituting the Origin $(0, 0)$

Substituting the Point $(4 a, 4 a)$

Simplifying the Relation using $a \neq = 0$

Constructing the Final Expression

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Dance of the Parabolas

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=0 and the line 2bx+3cy+4d=0 passes through the points of intersection of the parabolas y2=4ax and x2=4ay, then