If a hyperbola passes through the point P(10, 16) and it has vertices at (+-6, 0), then the equation

Q: Wait, why is the normal equation $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$? Can't I just find the slope of the tangent and take the negative reciprocal?

You absolutely can! That is the fundamental way to derive it. The tangent at $(x_1, y_1)$ is $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$. Its slope is $m_t = \frac{b^2 x_1}{a^2 y_1}$. The normal is perpendicular, so its slope is $m_n = -\frac{a^2 y_1}{b^2 x_1}$. Using the point-slope form $y - y_1 = m_n(x - x_1)$ will lead you to the exact same formula. Memorizing the formula just saves you time during the exam!

Q: How do I know if the hyperbola is horizontal or vertical?

Look at the vertices! If the vertices are on the x-axis, like $(\pm 6, 0)$, the transverse axis is horizontal. If they were on the y-axis, like $(0, \pm 6)$, it would be a vertical hyperbola. Always sketch a quick mental graph to orient yourself.

Q: Is there a risk of choosing the wrong sign for $b^2$?

In the standard hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $b^2$ is always a positive value representing the square of the semi-conjugate axis. If you ever calculate a negative value for $b^2$, double-check your arithmetic—you likely swapped a term or missed a negative sign during transposition.

Question

If a hyperbola passes through the point $P(10, 16)$ and it has vertices at $(\pm 6, 0)$, then the equation of the normal at $P$ is:

Options:

(A) $3x + 4y = 94$

(B) $x + 2y = 42$

(C) $2x + 5y = 100$

(D) $x + 3y = 58$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a hyperbola to find the equation of its normal.

We are given a hyperbola with vertices at $(\pm 6, 0)$. These vertices lie on the $x$-axis, which means our hyperbola is horizontal and centered perfectly at the origin $(0, 0)$.

The distance from the center to the vertex is the semi-transverse axis, denoted as $a$. Since the vertices are at $6$ and $-6$, we have $a = 6$, which implies $a^2 = 36$.

The standard equation for a horizontal hyperbola is:

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

Substituting our known value of $a^2$, we obtain:

$$\frac{x^2}{36} - \frac{y^2}{b^2} = 1$$

### Unlocking the Hidden Parameter

We are given that the point $P(10, 16)$ lies on the curve. Substituting $x = 10$ and $y = 16$ into our equation, we get:

$$\frac{10^2}{36} - \frac{16^2}{b^2} = 1$$

$$\frac{100}{36} - \frac{256}{b^2} = 1$$

Simplifying the fraction $\frac{100}{36}$ by dividing both the numerator and the denominator by $4$ gives $\frac{25}{9}$. Our equation becomes:

$$\frac{25}{9} - \frac{256}{b^2} = 1$$

Rearranging to isolate the term with $b^2$:

$$\frac{256}{b^2} = \frac{25}{9} - 1 = \frac{16}{9}$$

Solving for $b^2$ via cross-multiplication:

$$b^2 = \frac{256 	imes 9}{16} = 16 	imes 9 = 144$$

### The Normal Line Equation

The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at a point $(x_1, y_1)$ is given by the standard formula:

$$\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$$

Plugging in our values $a^2 = 36$, $b^2 = 144$, $x_1 = 10$, and $y_1 = 16$:

$$\frac{36x}{10} + \frac{144y}{16} = 36 + 144$$

### Final Calculation

Simplifying the coefficients, we note that $\frac{36}{10} = \frac{18}{5}$ and $\frac{144}{16} = 9$:

$$\frac{18x}{5} + 9y = 180$$

Multiplying the entire equation by $5$ to clear the fraction:

$$18x + 45y = 900$$

Dividing the entire equation by $9$ to reach the simplest form:

**$$2x + 5y = 100$$**

This is the final equation of the normal. You have navigated the geometry, conquered the algebra, and arrived at the truth.

Animated Solution for Mathematics - Conic Sections: If a hyperbola passes through the point $P (10, 16)$ and it has vertices at $(\pm 6, 0)$ , then the equation of the normal at $P$ is:

Select Answer:

Visualized Solution

Vertices of Hyperbola

Standard Equation

Point P on Hyperbola

Simplifying the Terms

Isolating $b^{2}$

Calculating $b^{2}$

Equation of Normal

Substituting Values

Simplifying the Equation

Final Equation of Normal

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Unlocking the Hidden Parameter

The Normal Line Equation

Final Calculation

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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; } If a hyperbola passes through the point P(10,16) and it has vertices at (±6,0), then the equation of the normal at P is: