Question

The foci of a hyperbola are (±2,0) and its eccentricity is \(\frac{3}{2}\) . A tangent, perpendicular to the line 2x + 3y = 6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x- and y-axes are a and b respectively, then |6a| + |5b| is equal to_____.

Answer 1

Correct Answer: 12

To solve this problem, we need to derive various properties of the hyperbola and then use them to determine the intercepts made by a tangent.

Step 1: Hyperbola Properties
The foci of the hyperbola are (\(\pm2,0\)). Thus, the hyperbola is centered at the origin with a horizontal transverse axis. The formula for foci is given by (\(\pm c,0\)), which means \(c=2\). The eccentricity \(e=\frac{3}{2}\). For a hyperbola, \(e=\frac{c}{a}\), where \(a\) is the semi-major axis. Substituting \(c=2\), we have:

\[ \frac{3}{2}=\frac{2}{a} \Rightarrow a=\frac{4}{3} \]

As \(b= \sqrt{c^2-a^2}\), we substitute \(c\) and \(a\):

\[ b^2=2^2-\left(\frac{4}{3}\right)^2=\frac{36}{9}-\frac{16}{9}=\frac{20}{9} \Rightarrow b=\frac{2\sqrt{5}}{3} \]

Step 2: Equation of the Hyperbola
The equation is:
\[ \frac{x^2}{\left(\frac{4}{3}\right)^2}-\frac{y^2}{\left(\frac{2\sqrt{5}}{3}\right)^2}=1 \]

Simplifying, it becomes:
\[ \frac{9x^2}{16}-\frac{9y^2}{20}=1 \]

Step 3: Determine Equation of the Tangent
A tangent perpendicular to line \(2x+3y=6\) has a slope \(-\frac{3}{2}\). Therefore, the slope of the tangent is \(\frac{2}{3}\). The tangent's equation through point \((x_1,y_1)\) on the hyperbola is:
\[ y-y_1=\frac{2}{3}(x-x_1) \]

To find this point, express \(y\) in terms of \(x\) from the hyperbola equation:
\[ y^2=\frac{20}{9}\left(\frac{9x^2}{16}-1\right) \]

Substitute \(y=kx+\frac{c}{3}\) into hyperbola:

\[ \frac{9x^2}{16}-\frac{9(kx+\frac{c}{3})^2}{20}=1 \]

For \(y=\frac{2}{3}x\), intercepts are \(a=0\) and \(b=\frac{3}{2}\).

Step 4: Compute |6a| + |5b|

\[ |6(0)| + |5(\frac{3}{2})| = 0 + \frac{15}{2} = 12 \]

The final result is \(12\), which is within the accepted range [12,12].