Which one of the following curves cut the parabola y² = 4ax at right angles? (A) x² + y² = a² (B) y

Q: Wait, why do we check the origin $(0,0)$? Is it the only point of intersection?

Great observation! In this specific problem, the curves $y^2 = 4ax$ and $x^2 = 4ay$ intersect at two points: $(0,0)$ and $(4a, 4a)$. We check the origin first because it is a critical point where the slopes become undefined for one curve and zero for the other, which is a classic case of orthogonality. You should always check all intersection points to be absolutely certain.

Q: What if the product of the slopes isn't exactly $-1$? Does that mean they aren't orthogonal?

Not necessarily! If the product of the slopes is $-1$, they are definitely orthogonal. However, if one slope is infinite (vertical tangent) and the other is zero (horizontal tangent), their product is technically undefined, but they are still perpendicular. Always look for this special case when dealing with vertical and horizontal tangents.

Q: Can I use implicit differentiation for all of these curves?

Absolutely! Implicit differentiation is often more robust than trying to solve for $y$ explicitly, especially when dealing with curves like $x^2 + y^2 = a^2$ or other conic sections. It keeps the algebra clean and helps you avoid potential errors with square roots.

Question

Which one of the following curves cut the parabola $y^2 = 4ax$ at right angles?

Options:

(A) $x^2 + y^2 = a^2$

(B) $y = e^{-x/2a}$

(C) $y = ax$

(D) $x^2 = 4ay$

Sigma Percentile · Accepted Answer

### The Geometry of Orthogonality: A Dance of Tangents

In the world of JEE Advanced, we don't just care that two curves meet; we care about *how* they meet. When we say two curves cut at right angles, or orthogonally, we are talking about the angle between their tangent lines at the point of intersection.

If that angle is exactly $90^{\circ}$, the curves are orthogonal. Mathematically, this is a beautiful condition: the product of their slopes, $m_1 \cdot m_2$, must be $-1$.

Alternatively, in the case where one tangent is vertical and the other is horizontal, they are inherently perpendicular.

### The First Step: Analyzing the Parabola

Let us start with our given curve, the parabola $y^2 = 4ax$. To find the slope of the tangent at any point, we differentiate both sides with respect to $x$:

$$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)$$

Using the chain rule on the left, we get $2y \frac{dy}{dx} = 4a$. Solving for the slope $m_1$, we find:

$$m_1 = \frac{dy}{dx} = \frac{2a}{y}$$

This expression provides the slope at any point $(x, y)$ on the parabola.

### The Hunt for the Candidate

Now, we test the candidate curve $x^2 = 4ay$. We apply the same logic by differentiating with respect to $x$:

$$2x = 4a \frac{dy}{dx}$$

This simplifies to the slope $m_2$:

$$m_2 = \frac{dy}{dx} = \frac{x}{2a}$$

We now have our two slope expressions: $m_1 = \frac{2a}{y}$ and $m_2 = \frac{x}{2a}$.

### The Intersection: Where Do They Meet?

Before we can check the orthogonality, we must find where these curves intersect. We solve $y^2 = 4ax$ and $x^2 = 4ay$ simultaneously.

Substituting $y = \frac{x^2}{4a}$ into the first equation, we get:

$$\left( \frac{x^2}{4a} ight)^2 = 4ax$$

This leads to $x^4 = 64a^3x$. Factoring this, we find $x(x^3 - 64a^3) = 0$. The solutions are $x = 0$ and $x = 4a$, yielding the intersection points $(0, 0)$ and $(4a, 4a)$.

### The Moment of Truth

Let us examine the slopes at the origin $(0, 0)$. For the first parabola, as $y 	o 0$, the slope $m_1 = \frac{2a}{y}$ approaches infinity, indicating a vertical tangent (the $y$-axis).

For the second parabola, at $x = 0$, the slope $m_2 = \frac{x}{2a}$ becomes $0$, indicating a horizontal tangent (the $x$-axis).

Since the $x$-axis and $y$-axis are perpendicular, the curves are orthogonal at the origin. By checking these slopes, we have confirmed that the curve **$x^2 = 4ay$** is indeed the one that cuts the parabola $y^2 = 4ax$ at right angles.

Animated Solution for Mathematics - Conic Sections: Which one of the following curves cut the parabola $y^{2} = 4 a x$ at right angles?

Select Answer:

Visualized Solution

The Orthogonality Condition

Slope of the Given Parabola

Calculating $m_{1}$

Testing the Candidate Curve

Calculating $m_{2}$

Finding Intersection Points

Roots of Intersection

Tangents at the Origin

Tangent of the Second Curve

Conclusion of Orthogonality

Final Answer

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

The Geometry of Orthogonality

The First Step

The Hunt for the Candidate

The Intersection

The Moment of Truth

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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; } Which one of the following curves cut the parabola y2=4ax at right angles?