Question

A hyperbola having transverse axis of length $\frac{1}{2}$ unit is confocal (having same foci) with the ellipse $3x^2+4y^2=12,$ then

• A. equation of the hyperbola is $x^2-\frac{y^2}{15}=\frac{1}{16}$
• B. eccenticity of the hyperbola is $4$
• C. distance between the directrices of the hyperbola is $\frac{1}{8}$ unit
• D. length of latus ractum of hyperbola is $\frac{15}{2}$ unit

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A equation of the hyperbola is $x^2-\frac{y^2}{15}=\frac{1}{16}$
B eccenticity of the hyperbola is $4$
C distance between the directrices of the hyperbola is $\frac{1}{8}$ unit
D length of latus ractum of hyperbola is $\frac{15}{2}$ unit

Equation of ellipse is
$3x^2+4y^2=12\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$
$\therefore e=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Foci are $(\pm 1,0)$

Now hyperbola is also having same focus i.e. $(\pm 1,0).$
If equation of hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $e_1$ is the eccentricity of hyberbola, then
$2a{e}_1=2$
Given $2a=\frac{1}{2}$
$\Rightarrow e_1=4$
and $b^2=a^2(e_1^2-1)=\frac{15}{16}$

So, the equation of the hyperbola is
$\frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{15}{16}}=1\Rightarrow \frac{x^2}{1}-\frac{y^2}{15}=\frac{1}{16}$

Distance between the directrices
$=\frac{2a}{e_1}=\frac{1}{8}$ unit

Length of latus ractum
$=\frac{2b^2}{a}=\frac{15}{2}$ unit