Statement-1 : An equation of a common tangent to the parabola y² = 16√3x and the ellipse 2x² + y² =

Q: Why do we use $y = mx + c$ instead of the point-slope form?

Great question! Using the point-slope form requires knowing the point of tangency, which we don't have yet. The slope-intercept form $y = mx + c$ is far more powerful here because it allows us to express the condition of tangency directly in terms of the slope $m$ and the parameters of the curves.

Q: Why did we square the equation? Doesn't that introduce extraneous roots?

Squaring is a classic algebraic maneuver to eliminate the square root. While it can introduce extraneous roots, we must always check our final values of $m$ against the original equation. In this case, since $m^2 = 4$ leads to real slopes $m = \pm 2$, our solutions are valid and physically meaningful.

Q: What if $m^2$ turned out to be negative? Would that mean no tangent exists?

Exactly! If the algebra led to $m^2 = -6$, it would imply that no real slope $m$ exists for a common tangent. This would geometrically mean the curves are positioned such that they don't share a common tangent. Always trust the algebra—it's the language of geometry!

Question

Statement-1 : An equation of a common tangent to the parabola $y^2 = 16\sqrt{3}x$ and the ellipse $2x^2 + y^2 = 4$ is $y = 2x + 2\sqrt{3}$ 
 Statement-2 : If the line $y = mx + \frac{4\sqrt{3}}{m}, (m 
eq 0)$ is a common tangent to the parabola $y^2 = 16\sqrt{3}x$ and the ellipse $2x^2 + y^2 = 4$, then $m$ satisfies $m^4 + 2m^2 = 24$

Options:

(A) Statement-1 is false, Statement-2 is true.

(B) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

(C) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

(D) Statement-1 is true, statement-2 is false.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two distinct geometric entities: a parabola $y^2 = 16\sqrt{3}x$ and an ellipse $2x^2 + y^2 = 4$. They seem like strangers, living in different parts of the plane, but they share a secret connection—a common tangent line.

Our mission today is to uncover this line and prove the relationship between these two curves.

### Standardizing the Ellipse

First, we must bring our ellipse into the light. The equation $2x^2 + y^2 = 4$ is not quite in its standard form.

To reveal its true nature, we divide the entire equation by $4$, yielding:

$$\frac{x^2}{2} + \frac{y^2}{4} = 1$$

Now, comparing this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we identify $a^2 = 2$ and $b^2 = 4$. This is the foundation of our ellipse.

### The Parabola's Identity

Next, we turn to the parabola $y^2 = 16\sqrt{3}x$. Comparing this to the standard form $y^2 = 4ax$, we find $4a = 16\sqrt{3}$, which simplifies beautifully to $a = 4\sqrt{3}$.

We now have the parameters for both curves.

### The Bridge of Tangency

A line $y = mx + c$ is a common tangent if it satisfies the condition of tangency for both curves. For the ellipse, the condition is $c^2 = a^2m^2 + b^2$.

Substituting our values, we get:

$$c^2 = 2m^2 + 4$$

For the parabola, the condition is $c = \frac{a}{m}$. Substituting $a = 4\sqrt{3}$, we get:

$$c = \frac{4\sqrt{3}}{m}$$

Now, we bridge the two worlds by equating the constant $c$. By substituting the parabola's $c$ into the ellipse's condition, we get:

$$\left( \frac{4\sqrt{3}}{m} ight)^2 = 2m^2 + 4$$

### The Algebraic Triumph

Now, let's solve this. Squaring the left side gives:

$$\frac{16 	imes 3}{m^2} = \frac{48}{m^2}$$

So, $\frac{48}{m^2} = 2m^2 + 4$. Multiplying by $m^2$ and rearranging, we arrive at:

$$2m^4 + 4m^2 - 48 = 0$$

Dividing by $2$, we get the elegant polynomial:

$$m^4 + 2m^2 - 24 = 0$$

This is the exact condition mentioned in Statement-2! Treating this as a quadratic in $m^2$, we factor it as $(m^2 + 6)(m^2 - 4) = 0$.

Since $m^2$ must be positive for a real slope, we find $m^2 = 4$, leading to $m = \pm 2$.

Substituting $m = 2$ back into our tangent equation $y = mx + \frac{4\sqrt{3}}{m}$, we get $y = 2x + \frac{4\sqrt{3}}{2}$, which simplifies to:

**$y = 2x + 2\sqrt{3}$**

This confirms Statement-1. We have successfully navigated the geometry and the algebra to find the common tangent. The beauty of this problem lies in how two seemingly different curves can be linked by a single, elegant line.

Select Answer:

Visualized Solution

Visualizing the Curves

Standardizing the Ellipse

Tangent Condition for Ellipse

Standardizing the Parabola

Tangent Condition for Parabola

Equating the Constants

Squaring Both Sides

Rearranging the Equation

Solving for Slope $m$

Verifying Statement 1

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Standardizing the Ellipse

The Parabola's Identity

The Bridge of Tangency

The Algebraic Triumph

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Curves

Standardizing the Ellipse

Tangent Condition for Ellipse

Standardizing the Parabola

Tangent Condition for Parabola

Equating the Constants

Squaring Both Sides

Rearranging the Equation

Solving for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Verifying Statement 1

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Standardizing the Ellipse

The Parabola's Identity

The Bridge of Tangency

The Algebraic Triumph

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