A hyperbola passes through the point P(√2, √3) and has foci at (±2, 0). Then the tangent to this hyp

Q: Wait, why do we reject $a^2 = 8$?

Excellent question! Remember that for a hyperbola, $b^2 = a^2(e^2 - 1)$. If $a^2 = 8$, then $b^2 = 4 - 8 = -4$, which is impossible for a real hyperbola. We need $b^2 > 0$, so $a^2$ must be less than 4.

Q: Why is the tangent equation $T=0$?

The notation $T=0$ is a powerful shorthand for the tangent at a point $(x_1, y_1)$ on a conic. For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, it replaces $x^2$ with $xx_1$ and $y^2$ with $yy_1$. It's a lifesaver in JEE exams!

Q: Can we use calculus to find the tangent?

Absolutely! You could differentiate $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with respect to $x$ to find the slope $\frac{dy}{dx}$. However, using $T=0$ is much faster and less prone to calculation errors under pressure.

Question

A hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the tangent to this hyperbola at P also passes through the point:

Options:

(A) $(3\sqrt{2}, 2\sqrt{3})$

(B) $(2\sqrt{2}, 3\sqrt{3})$

(C) $(\sqrt{3}, \sqrt{2})$

(D) $(-\sqrt{2}, -\sqrt{3})$

Sigma Percentile · Accepted Answer

### The Geometry of Balance

Imagine standing on the coordinate plane, looking at a hyperbola. It is not just a curve; it is a perfect balance of forces.

We are given a hyperbola that passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. The moment you see the foci at $(\pm 2, 0)$, you should immediately recognize that the transverse axis is the $x$-axis. This is the foundation of our journey.

### Decoding the Equation

We start with the standard equation of a hyperbola:

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

We know the foci are at $(\pm ae, 0)$. Comparing this to our given foci, we get $ae = 2$. Squaring this gives $a^2e^2 = 4$.

Now, recall the fundamental identity for a hyperbola: $b^2 = a^2(e^2 - 1)$. Expanding this, we get $b^2 = a^2e^2 - a^2$.

Substituting $a^2e^2 = 4$, we arrive at the elegant relation $b^2 = 4 - a^2$, or:

$$ a^2 + b^2 = 4 $$

This is the key that unlocks the entire problem.

### The Algebraic Dance

Since the point $P(\sqrt{2}, \sqrt{3})$ lies on the hyperbola, it must satisfy the equation. Substituting $x = \sqrt{2}$ and $y = \sqrt{3}$, we get:

$$ \frac{(\sqrt{2})^2}{a^2} - \frac{(\sqrt{3})^2}{b^2} = 1 \Rightarrow \frac{2}{a^2} - \frac{3}{b^2} = 1 $$

Now, we replace $b^2$ with $(4 - a^2)$. This gives us the equation:

$$ \frac{2}{a^2} - \frac{3}{4 - a^2} = 1 $$

Taking the LCM, we get:

$$ \frac{2(4 - a^2) - 3a^2}{a^2(4 - a^2)} = 1 $$

Expanding the numerator, we have $8 - 2a^2 - 3a^2 = 8 - 5a^2$. Setting this equal to the denominator $4a^2 - a^4$, we form the biquadratic equation:

$$ a^4 - 9a^2 + 8 = 0 $$

### The Final Reveal

Factorizing $a^4 - 9a^2 + 8 = 0$ is straightforward: $(a^2 - 1)(a^2 - 8) = 0$. This gives us $a^2 = 1$ or $a^2 = 8$.

But wait! We must respect the geometry. Since $b^2 = 4 - a^2$ and $b^2$ must be positive, $a^2$ must be less than 4.

Thus, we reject $a^2 = 8$ and accept $a^2 = 1$. Consequently, $b^2 = 3$. The equation of our hyperbola is:

$$ \frac{x^2}{1} - \frac{y^2}{3} = 1 $$

To find the tangent at $P(\sqrt{2}, \sqrt{3})$, we use the $T=0$ formula:

$$ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 $$

Substituting our values, we get:

$$ \frac{x\sqrt{2}}{1} - \frac{y\sqrt{3}}{3} = 1 $$

Multiplying by $\sqrt{3}$ simplifies this to $\sqrt{6}x - y = \sqrt{3}$.

Testing the point $(2\sqrt{2}, 3\sqrt{3})$, we find it satisfies this equation:

$$ \sqrt{6}(2\sqrt{2}) - 3\sqrt{3} = 2\sqrt{12} - 3\sqrt{3} = 4\sqrt{3} - 3\sqrt{3} = \sqrt{3} $$

The tangent passes through this point! You have successfully navigated the geometry and algebra of the hyperbola.

Animated Solution for Mathematics - Conic Sections: A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

Select Answer:

Visualized Solution

Visualizing the Given Data

Standard Equation of Hyperbola

Relating $a$ and $b$

Substituting Point $P$

Eliminating $b^{2}$

Forming the Quadratic Equation

Solving for $a^{2}$

The Hyperbola Equation

Equation of the Tangent

Substituting into Tangent Formula

Simplifying the Tangent Equation

Testing the Options

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Balance

Decoding the Equation

The Algebraic Dance

The Final Reveal

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