Question

The asymptotes of a hyperbola have equations $y-1=\frac{3}{4}(x+3).$ If a focus of the hyperbola has coordinates $(7,1)$, the equation of the hyperbola is

• A. $\frac{(x+3{)}^2}{16}-\frac{(y-1{)}^2}{9}=1$
• B. $\frac{(y-1{)}^2}{9}-\frac{(x+3{)}^2}{16}=1$
• C. $\frac{(x+3{)}^2}{64}-\frac{(y-1{)}^2}{36}=1$
• D. $\frac{(y-1{)}^2}{36}-\frac{(x+3{)}^2}{64}=1$
• E. $\frac{(x+3{)}^2}{4}-\frac{(y-1{)}^2}{3}=1$

Answer 1

Answer: (C)

$\frac{(x+3{)}^2}{64}-\frac{(y-1{)}^2}{36}=1$

Equation of asymptotes are

$y-1=\frac{3}{4}(x+3)$ ......(i)

$y-1=-\frac{3}{4}(x+3)$ .....(ii)

Centre of the hyperbola is point of intersection of asymptotes.

Therefore, by solving (i) and (ii), we get centre as $C(-3,1)$.

Slope of asymptotes $=\frac{b}{a}$

$\Rightarrow \frac{b}{a}=\pm \frac{3}{4}$ ......(i)

Focus is $(7,1)$.

Focus for hyperbola of form $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ is $(h+ae,k)$

$\Rightarrow 7=-3+ae\Rightarrow ae=10\Rightarrow a\frac{\sqrt{a^2+b^2}}{a}=10\Rightarrow \sqrt{a^2+b^2}=10$

Substituting $b$ from (i), we get

$\Rightarrow \sqrt{a^2+{(\pm a\frac{3}{4})}^2}=10\Rightarrow \frac{5a}{4}=10\Rightarrow a=8\Rightarrow b=\pm \frac{3}{4}a=\pm 6$

So, the equation of hyperbola is

$\frac{{(x+3)}^2}{8^2}-\frac{{(y-1)}^2}{6^2}=1$

$\frac{{(x+3)}^2}{64}-\frac{{(y-1)}^2}{36}=1$

So, option C is correct.