- Home
- Standard 11
- Mathematics
For the hyperbola $H : x ^{2}- y ^{2}=1$ and the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b>0$, let the
$(1)$ eccentricity of $E$ be reciprocal of the eccentricity of $H$, and
$(2)$ the line $y=\sqrt{\frac{5}{2}} x+K$ be a common tangent of $E$ and $H$ Then $4\left(a^{2}+b^{2}\right)$ is equal to
$2$
$0$
$1$
$3$
Solution
$e _{ E }=\sqrt{1-\frac{ b ^{2}}{ a ^{2}}}, e _{ H }=\sqrt{2}$
If $\Rightarrow e _{ E }=\frac{1}{ e _{ H }}$
$\Rightarrow \frac{ a ^{2}- b ^{2}}{ a ^{2}}=\frac{1}{2}$
$2 a ^{2-2 b } 2= a ^{2}$
$a ^{2}=2 b ^{2}$
and $y =\sqrt{\frac{5}{2}} x + k$ is tangent to ellipse then
$K ^{2}= a ^{2} \times \frac{5}{2}+ b ^{2}=\frac{3}{2}$
$\therefore b ^{2}=\frac{3}{2} \Rightarrow b ^{2}=\frac{1}{4}$ and $a ^{2}=\frac{1}{2}$
$\left.\therefore a ^{2}+ b ^{2}\right)=3$
