If 2x - y + 1 = 0 is a tangent to the hyperbola x²/a² - y²/16 = 1, then which of the following CANNO

Q: Why is the condition $c^2 = A^2m^2 - B^2$ and not $c^2 = A^2m^2 + B^2$?

That is a fantastic question! That plus sign belongs to the ellipse. For a hyperbola, the geometry is 'opening out,' which changes the sign in the tangency condition. Always keep your conic section formulas distinct by visualizing the curve!

Q: How do I know which side is the hypotenuse in the options?

Always look for the largest value. Since $\sqrt{17} \approx 4.12$, you can compare the magnitudes of the sides in each option to identify the candidate for the hypotenuse. For example, in option A, $\sqrt{17}$ is clearly the largest.

Q: The question asks for 'CANNOT' be sides. Does that mean I need to find the ones that fail?

Exactly! This is a classic JEE trap. You are looking for the options that do NOT satisfy the Pythagorean theorem. Don't stop at the first correct one; check all of them carefully.

Question

If $2x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$, then which of the following CANNOT be sides of a right angled triangle ?

Options:

(A) $2a, 4, 1$

(B) $2a, 8, 1$

(C) $a, 4, 1$

(D) $a, 4, 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we embark on a journey through the elegant world of coordinate geometry. We are given a hyperbola and a line, and we are told that this line is a tangent.

The hyperbola is defined by:

$$
\frac{x^2}{a^2} - \frac{y^2}{16} = 1
$$

Our line is $2x - y + 1 = 0$. To make this line speak to us, let us rearrange it into the slope-intercept form: $y = 2x + 1$. Here, the slope $m$ is $2$, and the intercept $c$ is $1$.

### The Condition of Tangency

Now, we must invoke the condition of tangency. For a hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, the line $y = mx + c$ is tangent if and only if:

$$
c^2 = A^2m^2 - B^2
$$

This formula is the key that unlocks the door. Substituting our values, we have:

$$
1^2 = a^2(2)^2 - 16
$$

This simplifies beautifully to $1 = 4a^2 - 16$. Adding $16$ to both sides, we get $17 = 4a^2$, which means $a^2 = \frac{17}{4}$.

Taking the square root, we find $a = \frac{\sqrt{17}}{2}$, and consequently, $2a = \sqrt{17}$.

### The Pythagorean Test

We are now ready to test the options. The question asks us to identify which sets of sides **CANNOT** form a right-angled triangle. We must use the Pythagorean theorem: $p^2 + q^2 = r^2$, where $r$ is the longest side.

Let us test Option A: $2a, 4, 1$. Substituting $2a = \sqrt{17}$, we get $\sqrt{17}, 4, 1$. The squares are $17, 16, 1$. Since $1 + 16 = 17$, this set forms a right-angled triangle.

Now, Option B: $2a, 8, 1$. We have $\sqrt{17}, 8, 1$. The squares are $17, 64, 1$. The sum of the smaller squares is $1 + 17 = 18$, which is not $64$. This fails!

Option C: $a, 4, 1$. We have $\frac{\sqrt{17}}{2}, 4, 1$. The squares are $\frac{17}{4}, 16, 1$. The sum of the smaller squares is:

$$
1 + \frac{17}{4} = \frac{21}{4} 
eq 16
$$

This also fails!

Finally, Option D: $a, 4, 2$. We have $\frac{\sqrt{17}}{2}, 4, 2$. The squares are $\frac{17}{4}, 16, 4$. The sum of the smaller squares is:

$$
4 + \frac{17}{4} = \frac{33}{4} 
eq 16
$$

This fails as well!

### Conclusion

The beauty of this problem lies in the trap. We were looking for the sets that **CANNOT** form a triangle.

By systematically applying the Pythagorean theorem, we discovered that options **B, C, and D** are the correct choices. Keep practicing, keep visualizing, and never let the algebra intimidate you.

Animated Solution for Mathematics - Conic Sections: If $2 x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{16} = 1$ , then which of the following CANNOT be sides of a right angled triangle ?

Select Answer:

Visualized Solution

The Hyperbola & Tangent

Condition of Tangency

Substituting Values

Solving for $a^{2}$

Finding $a$ and $2 a$

The Right-Angled Triangle Condition

Checking Option A

Checking Option B

Checking Option C

Checking Option D

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Condition of Tangency

The Pythagorean Test

Conclusion

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