The number of values of c such that the straight line y = 4x + c touches the curve (x²/4) + y² = 1 i

Q: Wait, why do we use the formula $c^2 = a^2m^2 + b^2$ instead of setting the discriminant to zero?

You absolutely can use the discriminant method! If you substitute $y = 4x + c$ into the ellipse equation and set the discriminant of the resulting quadratic to zero, you will get the exact same result. The formula $c^2 = a^2m^2 + b^2$ is simply a derived shortcut that saves time in competitive exams like JEE. Both methods are mathematically sound.

Q: Why are there two values of $c$? Does that mean there are two ellipses?

Not at all! There is only one ellipse. The two values of $c$ correspond to the two parallel tangent lines that can be drawn to the ellipse for a given slope. Imagine the ellipse centered at the origin; one tangent touches the 'top-right' side, and the other touches the 'bottom-left' side. They are parallel to each other, hence they share the same slope $m=4$ but have different intercepts $c$.

Q: What if the line was given as $x = 4y + c$? Would the formula still work?

Great question! The formula $c^2 = a^2m^2 + b^2$ is specific to the form $y = mx + c$. If your line is $x = 4y + c$, you must first rearrange it into the slope-intercept form: $y = \frac{1}{4}x - \frac{c}{4}$. Now, your slope $m$ is $\frac{1}{4}$ and your intercept is $-\frac{c}{4}$. Always ensure your equation is in the correct form before applying standard formulas.

Question

The number of values of $c$ such that the straight line $y = 4x + c$ touches the curve $(x^2/4) + y^2 = 1$ is

Options:

(A) 0

(B) 1

(C) 2

(D) infinite

Sigma Percentile · Accepted Answer

### The Geometry of Tangency: A Journey into Coordinate Elegance

Welcome, fellow traveler of the mathematical landscape. Today, we are going to unravel a classic problem in coordinate geometry.

It is not just about finding an answer; it is about seeing the hidden symmetry of the ellipse and the elegant dance of lines across the Cartesian plane. Let us begin.

### Phase 1: The Ellipse as Our Stage

We start with the curve:

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

This is the standard equation of an ellipse, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. By comparing the two, we immediately identify our parameters: $a^2 = 4$ and $b^2 = 1$.

Imagine this ellipse centered at the origin, stretched along the $x$-axis. It is our stage, and we are about to introduce an actor.

### Phase 2: The Sliding Line

Our actor is the line $y = 4x + c$. Notice something crucial here: the slope $m = 4$ is fixed.

This means the line has a steep, constant inclination. However, the $y$-intercept $c$ is a variable.

As $c$ changes, the line slides up and down, remaining perfectly parallel to itself. We are looking for the specific values of $c$ where this line 'touches' the ellipse—a state we call **tangency**.

### Phase 3: The Power of the Tangency Condition

In the world of JEE, we love tools that simplify complexity. For a line $y = mx + c$ to be tangent to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, there is a beautiful, ironclad condition:

$$c^2 = a^2m^2 + b^2$$

Why does this work? If you were to substitute the line equation into the ellipse equation, you would get a quadratic in $x$.

For the line to touch the curve at exactly one point, that quadratic must have a discriminant of zero. This formula is the distilled essence of that requirement.

### Phase 4: The Calculation

Now, let us bring our parameters into the light. We have $a^2 = 4$, $b^2 = 1$, and $m = 4$. Substituting these into our condition, we get:

$$c^2 = (4)(4)^2 + 1$$

Take a breath. Let us calculate this carefully. $4^2$ is $16$. Multiplying by $4$ gives us $64$. Adding $1$ brings us to $65$.

So, we have:

$$c^2 = 65$$

### Phase 5: The Final Revelation

We are almost there. To find $c$, we take the square root of both sides. Remember, when we solve $c^2 = 65$, we must account for both the positive and negative roots:

$$c = \pm\sqrt{65}$$

This gives us two distinct values: $c = \sqrt{65}$ and $c = -\sqrt{65}$. These correspond to the two parallel tangent lines, one on either side of the ellipse.

Thus, there are exactly **two values of $c$** that satisfy our condition. We have successfully navigated the geometry, applied the algebraic tool, and arrived at the truth.

Keep this intuition with you: for any given slope, an ellipse will generally have two parallel tangents. You have mastered the logic!

Animated Solution for Mathematics - Conic Sections: The number of values of $c$ such that the straight line $y = 4 x + c$ touches the curve $(x^{2} /4) + y^{2} = 1$ is

Select Answer:

Visualized Solution

Visualizing the Ellipse

The Family of Lines

Condition for Tangency

Identifying Parameters

Substituting the Values

Evaluating $c^{2}$

Solving for $c$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Tangency

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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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; } The number of values of c such that the straight line y=4x+c touches the curve (x2/4)+y2=1 is