The foci of a hyperbola are [( 2,0)] and its eccentricity is [(3)/(2)].

Question

The foci of a hyperbola are $$( \pm 2,0)$$ and its eccentricity is $$\frac{3}{2}$$. A tangent, perpendicular to the line $2 x+3 y=6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the...

Answer: 12

Accepted Answer

Given that the foci are at $(\pm 2,0)$, the distance between the foci is $2ae=2	imes 2=4$.

So, $a=\frac{4}{3}$.

The eccentricity of the hyperbola is given as $\frac{3}{2}$. Thus, $e=\frac{3}{2}$.

For a hyperbola, ${e}^{2}=1+\frac{{b}^{2}}{{a}^{2}}$, which gives ${b}^{2}={a}^{2}({e}^{2}-1)$.

Substituting the given values, we get ${b}^{2}=\frac{16}{9}-\frac{16}{9}=\frac{20}{9}$.

The equation of the hyperbola is $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$, or $\frac{9{x}^{2}}{16}-\frac{9{y}^{2}}{20}=1$.

The slope of the tangent line, which is perpendicular to the given line $2x+3y=6$, is $\frac{3}{2}$.

The equation of the tangent to a hyperbola is $y=mx\pm \sqrt{{a}^{2}{m}^{2}-{b}^{2}}$. Substituting the given values, we get $y=\frac{3x}{2}\pm \sqrt{\frac{16}{9}	imes \frac{9}{4}-\frac{20}{9}}=\frac{3x}{2}\pm \frac{4}{3}$.

The $x$-intercept occurs when $y=0$, so $|a|=\frac{8}{9}$, and the $y$-intercept occurs when $x=0$, so $|b|=\frac{4}{3}$.

Finally, $|6a|+|5b|=6\frac{8}{9}+5\frac{4}{3}=\frac{48}{9}+\frac{60}{9}=\frac{108}{9}=12$.

Solutions · 1