The point P(-2√6, √3) lies on the hyperbola x²/a² - y²/b² = 1 having eccentricity √5/2. If the tange

Q: Wait, why do we use the conjugate axis for the intersection?

The conjugate axis of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the $y$-axis, which corresponds to the line $x=0$. The problem asks for the intersection with this axis, so setting $x=0$ is the direct and correct path to find the intercepts.

Q: Can I use the parametric form for the tangent and normal?

Absolutely! You could use $x = a \sec \theta$ and $y = b \tan \theta$. However, since you are already given the specific coordinates of point $P$, the point-form of the tangent $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$ is much faster and less prone to algebraic errors.

Q: Is the normal slope always the negative reciprocal of the tangent slope?

Yes, for any differentiable curve, the normal line is defined as being perpendicular to the tangent line at the point of contact. If the tangent has slope $m_T$, the normal must have slope $m_N = -\frac{1}{m_T}$. This is a fundamental property of geometry that holds true here.

Question

The point $P(-2\sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ having eccentricity $\frac{\sqrt{5}}{2}$. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :

Options:

(A) $4\sqrt{3}$

(B) 6

(C) $6\sqrt{3}$

(D) $3\sqrt{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of a hyperbola is defined by the relationship between its semi-axes $a$ and $b$ and its eccentricity $e$. We are given the eccentricity $e = \frac{\sqrt{5}}{2}$.

In the study of conic sections, the fundamental relation is $b^2 = a^2(e^2 - 1)$. Substituting the given eccentricity, we obtain:

$$b^2 = a^2 \left( \left( \frac{\sqrt{5}}{2} \right)^2 - 1 \right)$$

Squaring the eccentricity yields $\frac{5}{4}$, and subtracting $1$ results in $\frac{1}{4}$. Thus, we arrive at the relation:

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

### The Point of Contact

Point $P(-2\sqrt{6}, \sqrt{3})$ lies on the hyperbola, so it must satisfy the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substituting the coordinates of $P$, we get:

$$\frac{(-2\sqrt{6})^2}{a^2} - \frac{(\sqrt{3})^2}{b^2} = 1 \Rightarrow \frac{24}{a^2} - \frac{3}{b^2} = 1$$

Using our earlier discovery that $a^2 = 4b^2$, we substitute to solve for $b^2$:

$$\frac{24}{4b^2} - \frac{3}{b^2} = 1 \Rightarrow \frac{6}{b^2} - \frac{3}{b^2} = 1$$

This simplifies to $\frac{3}{b^2} = 1$, which gives $b^2 = 3$ and $a^2 = 12$. The equation of our hyperbola is:

$$\frac{x^2}{12} - \frac{y^2}{3} = 1$$

### The Tangent's Dance

The equation for a tangent at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. Plugging in our values, we obtain:

$$\frac{x(-2\sqrt{6})}{12} - \frac{y(\sqrt{3})}{3} = 1 \Rightarrow -\frac{x}{\sqrt{6}} - \frac{y}{\sqrt{3}} = 1$$

To find point $Q$, where this tangent intersects the conjugate axis, we set $x=0$:

$$-\frac{y}{\sqrt{3}} = 1 \Rightarrow y = -\sqrt{3}$$

Thus, the coordinates of $Q$ are $(0, -\sqrt{3})$.

### The Normal's Path

The normal is perpendicular to the tangent at $P$. Rearranging the tangent equation into slope-intercept form $y = mx + c$:

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

The slope of the tangent is $m_T = -\frac{1}{\sqrt{2}}$. Therefore, the slope of the normal is $m_N = \sqrt{2}$. Using the point-slope form $y - y_1 = m_N(x - x_1)$:

$$y - \sqrt{3} = \sqrt{2}(x + 2\sqrt{6})$$

To find point $R$, the intersection with the conjugate axis, we set $x=0$:

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

Solving for $y$, we find $y = 5\sqrt{3}$. Thus, the coordinates of $R$ are $(0, 5\sqrt{3})$.

### Final Calculation

We have identified the points $Q(0, -\sqrt{3})$ and $R(0, 5\sqrt{3})$. Since both points lie on the $y$-axis, the distance $QR$ is the absolute difference of their $y$-coordinates:

$$QR = |5\sqrt{3} - (-\sqrt{3})| = 6\sqrt{3}$$

The final distance is **$6\sqrt{3}$**.

Select Answer:

Visualized Solution

Visualizing the Problem

Relating $a^{2}$ and $b^{2}$

Simplifying the Relation

Using Point $P$

Solving for $a^{2}$ and $b^{2}$

Equation of Tangent at $P$

Locating Point $Q$

Slope of Normal

Equation of Normal at $P$

Locating Point $R$

Calculating the Length $QR$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Point of Contact

The Tangent's Dance

The Normal's Path

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Problem

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Simplifying the Relation

Using Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q