Let A be a point on the x-axis. Common tangents are drawn from A to the curves x² + y² = 8 and y² =

Q: Wait, why do we use the slope form $y = mx + \frac{a}{m}$ instead of the point form?

Great question! Using the slope form is a strategic shortcut. Since the tangent is common to both, it must have a single slope $m$ that satisfies the tangency condition for both curves. By using $m$ as our variable, we reduce the problem to a single algebraic equation, which is much cleaner than dealing with two separate points of contact.

Q: Why did we get $m^2 = 1$ and reject $m^2 = -2$?

In the real coordinate plane, the slope $m$ must be a real number. Since $m^2 = -2$ would imply $m = \pm i\sqrt{2}$, which is imaginary, we discard it. We only care about real tangents that exist on the Cartesian plane!

Q: Is there a specific order to find the points $Q$ and $R$?

Not strictly, but it is mathematically safer to find the parabola's contact point using the standard formula $(\frac{a}{m^2}, \frac{2a}{m})$ and the circle's point by substituting the line equation into the circle's equation. This ensures you don't miss any subtle signs.

Q: Why does the circle equation result in $(x+2)^2 = 0$?

That is the beauty of tangency! When you substitute the tangent line into the circle's equation, you get a quadratic in $x$. Because the line is tangent, it touches at exactly one point, meaning the quadratic must be a perfect square. If you don't get a perfect square, it's a sign to recheck your algebra.

Q: Can I use the distance formula between the two points directly?

Absolutely! Once you have the coordinates of $Q$ and $R$, the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ is the final step. Just be careful with the signs when subtracting negative coordinates.

Question

Let A be a point on the x-axis. Common tangents are drawn from A to the curves $x^2 + y^2 = 8$ and $y^2 = 16x$. If one of these tangents touches the two curves at Q and R, then $(QR)^2$ is equal to

Options:

(A) 64

(B) 76

(C) 81

(D) 72

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we aren't just solving a coordinate geometry problem; we are uncovering the hidden harmony between two fundamental shapes: the circle and the parabola.

Imagine standing on the Cartesian plane. You have a circle, $x^2 + y^2 = 8$, a perfect, symmetric loop centered at the origin with a radius of $r = 2\sqrt{2}$.

Beside it, a parabola $y^2 = 16x$ stretches out, a curve of infinite potential. Our goal is to find the common tangent—a bridge that touches both curves—and calculate the distance between these two points of contact, $Q$ and $R$.

### The Power of the Slope

When faced with a common tangent, the most powerful tool in your arsenal is the slope-intercept form. For a parabola $y^2 = 4ax$, any tangent can be described by the elegant equation $y = mx + \frac{a}{m}$.

Here, our parabola $y^2 = 16x$ gives us $4a = 16$, meaning $a = 4$. Thus, our tangent line is $y = mx + \frac{4}{m}$.

Think of $m$ as the 'DNA' of this line. If we find the right $m$, we define the line completely. We rearrange this into the standard form: $mx - y + \frac{4}{m} = 0$.

### The Condition of Tangency

How do we ensure this line touches the circle? Geometry provides a beautiful constraint: the perpendicular distance from the center of the circle $(0,0)$ to the line must be exactly equal to the radius $r = 2\sqrt{2}$.

Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we set up the equation:

$$\frac{|\frac{4}{m}|}{\sqrt{m^2 + (-1)^2}} = 2\sqrt{2}$$

Squaring both sides is our next logical step to clear the radical and the absolute value. This leads us to the following expression:

$$\frac{16}{m^2(m^2 + 1)} = 8$$

Simplifying this, we arrive at the biquadratic equation $m^4 + m^2 - 2 = 0$.

### Solving the Mystery

This looks intimidating, but look closer! It is a quadratic in disguise. Let $u = m^2$.

Then $u^2 + u - 2 = 0$. Factoring this, we get $(u + 2)(u - 1) = 0$.

Since $m^2$ cannot be $-2$ in the real plane, we must have $m^2 = 1$, which gives us $m = \pm 1$. Let's choose $m = 1$ for our calculation. Our tangent line is now simply $y = x + 4$.

### Finding the Points of Contact

Now, we locate the points $Q$ and $R$. For the parabola, the point of contact is given by the formula $(\frac{a}{m^2}, \frac{2a}{m})$.

Substituting $a = 4$ and $m = 1$, we find $R = (4, 8)$.

For the circle, we substitute $y = x + 4$ into $x^2 + y^2 = 8$. This yields $x^2 + (x + 4)^2 = 8$, which simplifies to $2x^2 + 8x + 8 = 0$.

Dividing by $2$, we get $(x + 2)^2 = 0$. This confirms our tangency! The point $Q$ is at $x = -2$, and since $y = x + 4$, $y = 2$. So, $Q = (-2, 2)$.

### Final Calculation

We have our points: $Q(-2, 2)$ and $R(4, 8)$. The final step is to calculate the square of the distance between them:

$$(QR)^2 = (4 - (-2))^2 + (8 - 2)^2$$

$$(QR)^2 = 6^2 + 6^2 = 36 + 36 = 72$$

And there it is. The elegance of the result, **72**, is the reward for your persistence. You didn't just calculate a number; you navigated the intersection of two distinct geometric worlds.

Animated Solution for Mathematics - Conic Sections: Let A be a point on the x-axis. Common tangents are drawn from A to the curves $x^{2} + y^{2} = 8$ and $y^{2} = 16 x$ . If one of these tangents touches the two curves at Q and R, then $(QR)^{2}$ is equal to

Select Answer:

Visualized Solution

Visualize the Geometry

Tangent to the Parabola

Tangent Equation Setup

Condition for Circle Tangency

Applying the Distance Formula

Solving for Slope $m$

Finding the Value of $m$

Finding Contact Point $R$

Coordinates of Point $R$

Finding Contact Point $Q$

Coordinates of Point $Q$

Calculate $(QR)^{2}$

Final Computation

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of the Slope

The Condition of Tangency

Solving the Mystery

Finding the Points of Contact

Final Calculation

Similar Questions

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Let the lines $y + 2 x = 11 + 77$ and $2 y + x = 211 + 67$ be normal to a circle $C : (x - h)^{2} + (y - k)^{2} = r^{2}$ . If the line $11 y - 3 x = \frac{5 77}{3} + 11$ is tangent to the circle $C$ , then the value of $(5 h - 8 k)^{2} + 5 r^{2}$ is equal to ____.

Select Answer:

Visualized Solution

Visualize the Geometry

Tangent to the Parabola

Tangent Equation Setup

Condition for Circle Tangency

Applying the Distance Formula

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