Question

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 $ If the eccentricity of the hyperbola is 2, then the equation of the tangent to this hyperbola passing through the point (4, 6) is :

• A. 3x - 2y = 0
• B. 2x - 3y + 10 = 0
• C. x - 2y + 8 = 0
• D. 2x - y - 2 = 0

Answer 1

The Correct Option is C

 To solve the problem, we first need to understand the position of the foci of both the ellipse and the hyperbola. Given that the foci of the hyperbola coincide with those of the ellipse, we start by finding the foci of the given ellipse.

The equation of the ellipse is:

\(\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1\)

This is a standard form of an ellipse equation where \(a^2 = 25\) and \(b^2 = 9\). Here, \(a = 5\) and \(b = 3\).

The formula for the focal distance of an ellipse (distance from the center to each focus) is:

\(c = \sqrt{a^2 - b^2}\)  =>  \(c = \sqrt{25 - 9} = \sqrt{16} = 4\)

Thus, the foci of the ellipse are \((\pm 4, 0)\).

Since the foci of the hyperbola coincide with these, the hyperbola also has foci at \((\pm 4, 0)\).

Given the eccentricity (e) of the hyperbola is 2, we use the formula for the eccentricity of a hyperbola: \(e = \frac{c}{a}\). Here, \(c = 4\).

Thus, \(2 = \frac{4}{a}\) implies \(a = 2\).

The relationship between a, b, and c in a hyperbola is \(c^2 = a^2 + b^2\). Since \(c = 4\) and \(a = 2\):

\(4^2 = 2^2 + b^2\)  =>  \(16 = 4 + b^2\)  =>  \(b^2 = 12\)

The equation of the hyperbola is then:

\(\frac{x^2}{4} - \frac{y^2}{12} = 1\)

To find the equation of the tangent to this hyperbola passing through the point (4, 6), we use the point-slope form of the tangent to a hyperbola:

\(\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1\) where \((x_1, y_1)\) is (4, 6).

This becomes \(\frac{x \cdot 4}{4} - \frac{y \cdot 6}{12} = 1\).

Simplifying: \(x - \frac{y}{2} = 1\)  =>  multiply throughout by 2 gives \(2x - y = 2\).

To match the given options, let's rearrange: \(x - 2y + 8 = 0\). On closer examination, this matches with one of the provided options.

Therefore, the equation of the tangent to the hyperbola passing through (4, 6) is \(x - 2y + 8 = 0\).