Question

If \(3x+2\sqrt{2}y+k=0\) is a normal to the hyperbola \(4x^2-9y^2-36=0\) making positive intercepts on both the axes, then \(k=\) Options :

• A. \(13\sqrt{2}\)
• B. \(-5\sqrt{2}\)
• C. \(-2\sqrt{2}\)
• D. \(-13\sqrt{2}\)

Hint:

To find the equation of a normal to a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) at a point \((x_1, y_1)\), use the formula \(\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2\). If a given line is a normal, compare its coefficients with this general form to find \(x_1, y_1\) in terms of the unknown parameter. Then, substitute \((x_1, y_1)\) into the hyperbola's equation to solve for the parameter. Finally, use any additional conditions (like intercept signs) to determine the specific value of the parameter.

Answer 1

The Correct Option is D

Step 1: Convert the hyperbola equation to standard form.
The given equation of the hyperbola is \(4x^2 - 9y^2 - 36 = 0\).
Rearrange to the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\):
\[ 4x^2 - 9y^2 = 36 \] Divide by 36: \[ \frac{4x^2}{36} - \frac{9y^2}{36} = 1 \] \[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \] From this, we have \(a^2 = 9 \implies a = 3\) and \(b^2 = 4 \implies b = 2\).
Step 2: Use the condition for a normal to a hyperbola.
The equation of a 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:
\[ \frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2 \] Substitute \(a^2 = 9\) and \(b^2 = 4\): \[ \frac{9x}{x_1} + \frac{4y}{y_1} = 9 + 4 \] \[ \frac{9x}{x_1} + \frac{4y}{y_1} = 13 \quad \text{(Equation of Normal)} \] The given normal line is \(3x + 2\sqrt{2}y + k = 0\), which can be written as \(3x + 2\sqrt{2}y = -k\).
Comparing the coefficients of the two normal equations: \[ \frac{9/x_1}{3} = \frac{4/y_1}{2\sqrt{2}} = \frac{13}{-k} \] \[ \frac{3}{x_1} = \frac{2}{\sqrt{2}y_1} = \frac{13}{-k} \] From the first equality: \[ \frac{3}{x_1} = \frac{2}{\sqrt{2}y_1} \implies \frac{3}{x_1} = \frac{\sqrt{2}}{y_1} \implies y_1 = \frac{\sqrt{2}}{3}x_1 \quad \text{(Relation between \(x_1\), \(y_1\))} \] From the second equality: \[ \frac{3}{x_1} = \frac{13}{-k} \implies x_1 = -\frac{3k}{13} \] \[ \frac{\sqrt{2}}{y_1} = \frac{13}{-k} \implies y_1 = -\frac{\sqrt{2}k}{13} \] Step 3: Use the condition that \((x_1, y_1)\) lies on the hyperbola.
Since \((x_1, y_1)\) is a point on the hyperbola \(\frac{x^2}{9} - \frac{y^2}{4} = 1\), it must satisfy its equation: \[ \frac{x_1^2}{9} - \frac{y_1^2}{4} = 1 \] Substitute the expressions for \(x_1\) and \(y_1\) in terms of \(k\): \[ \frac{\left(-\frac{3k}{13}\right)^2}{9} - \frac{\left(-\frac{\sqrt{2}k}{13}\right)^2}{4} = 1 \] \[ \frac{9k^2/169}{9} - \frac{2k^2/169}{4} = 1 \] \[ \frac{k^2}{169} - \frac{2k^2}{169 \times 4} = 1 \] \[ \frac{k^2}{169} - \frac{k^2}{338} = 1 \] Find a common denominator: \[ \frac{2k^2 - k^2}{338} = 1 \] \[ \frac{k^2}{338} = 1 \implies k^2 = 338 \] So, \(k = \pm\sqrt{338} = \pm\sqrt{169 \times 2} = \pm 13\sqrt{2}\).
Step 4: Use the condition that the normal makes positive intercepts on both axes.
The equation of the normal is \(3x + 2\sqrt{2}y + k = 0\).
To find the x-intercept, set \(y = 0\): \(3x + k = 0 \implies x = -\frac{k}{3}\).
To find the y-intercept, set \(x = 0\): \(2\sqrt{2}y + k = 0 \implies y = -\frac{k}{2\sqrt{2}}\).
For both intercepts to be positive, we must have: \(-\frac{k}{3} > 0 \implies k < 0\)
\(-\frac{k}{2\sqrt{2}} > 0 \implies k < 0\)
Both conditions require \(k\) to be negative.
From \(k = \pm 13\sqrt{2}\), the value that satisfies \(k < 0\) is \(k = -13\sqrt{2}\).