Question

The normal to the hyperbola
\(\frac{x²}{a²} - \frac{y²}{9} = 1\)
at the point (8, 3√3) on it passes through the point:

• A. \((15, -2\sqrt3)\)
• B. \((9, 2\sqrt3)\)
• C. \(( -1, 9\sqrt3)\)
• D. \(( -1, 6\sqrt3)\)

Answer 1

The Correct Option is C

To find the point through which the normal at a given point on the hyperbola passes, we start by examining the equation of the hyperbola and determining the necessary parameters to derive the equation of the normal.

The standard form of the hyperbola in question is:

\(\frac{x^2}{a^2} - \frac{y^2}{9} = 1\)

From this equation, we know:

• The center of the hyperbola is at the origin (0, 0).
• The point (8, 3√3) is given to lie on the hyperbola, helping determine a^2 as follows:

Substituting \((x, y) = (8, 3\sqrt{3})\) into the equation gives:

\(\frac{8^2}{a^2} - \frac{(3\sqrt{3})^2}{9} = 1\)

Solving this,

\(\frac{64}{a^2} - \frac{27}{9} = 1\) \(\frac{64}{a^2} - 3 = 1\) \(\frac{64}{a^2} = 4\) a^2 = 16

Thus, the hyperbola is:

\(\frac{x^2}{16} - \frac{y^2}{9} = 1\)

The slope of the tangent at any point (x_1, y_1) on the hyperbola is given by:

\(\frac{y_1}{b^2} = \frac{y_1}{9}\) \(\frac{x_1}{a^2} = \frac{x_1}{16}\) Slope = \frac{b^2 \cdot x_1}{a^2 \cdot y_1} = \frac{9 \cdot x_1}{16 \cdot y_1}

For point \((8, 3\sqrt{3})\), the slope of the tangent is:

\(\frac{9 \cdot 8}{16 \cdot 3\sqrt{3}} = \frac{72}{48\sqrt{3}} = \frac{3}{2\sqrt{3}}\)

Simplifying gives a slope of the normal:

-\frac{dy}{dx} = \frac{3}{2\sqrt{3}} \Rightarrow Slope\: of\: Normal = -\frac{2\sqrt{3}}{3}

The equation of the normal line at the point \((8, 3\sqrt{3})\) with slope m is:

y - y_1 = m(x - x_1)

Substituting the slope and point:

y - 3\sqrt{3} = -\frac{2\sqrt{3}}{3}(x - 8)

Expanding and rearranging,

y = -\frac{2\sqrt{3}}{3}x + \frac{16\sqrt{3} + 9\sqrt{3}}{3} y = -\frac{2\sqrt{3}}{3}x + 9\sqrt{3}

To find if this line passes through any of the given points, check each option:

• Option (-1, 9\sqrt{3}): 9\sqrt{3} = -\frac{2\sqrt{3}}{3}(-1) + 9\sqrt{3} 9\sqrt{3} = \frac{2\sqrt{3}}{3} + 9\sqrt{3} The point satisfies the normal line equation.

Therefore, the correct answer is that the normal passes through the point:

(-1, 9\sqrt{3})