Consider a branch of hyperbola x^2-2 y^2-2 √{2} x-4 √{2} y-6=0 with vertex at point A . Let B be

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10-2. Parabola, Ellipse, Hyperbola

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Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $\mathrm{A}$, then the area of the triangle $\mathrm{ABC}$ is

$1-\sqrt{\frac{2}{3}}$

$\sqrt{\frac{3}{2}}-1$

$1+\sqrt{\frac{2}{3}}$

$\sqrt{\frac{3}{2}}+1$

(IIT-2008)

Solution

Hyperbola is $\frac{(x-\sqrt{2})^2}{4}-\frac{(y+\sqrt{2})^2}{2}=1$

$ a=2, b=\sqrt{2} $

$ e=\sqrt{\frac{3}{2}}$

Area $=\frac{1}{2} \mathrm{a}(\mathrm{e}-1) \times \frac{\mathrm{b}^2}{\mathrm{a}}=\frac{1}{2} \frac{(\sqrt{3}-\sqrt{2}) \times 2}{\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})}{\sqrt{2}}$

$\Rightarrow \text { Area }=\left(\sqrt{\frac{3}{2}}-1\right)$

Standard 11

Mathematics

The equation of the tangent to the hyperbola $4{y^2} = {x^2} – 1$ at the point $(1, 0)$ is

easy

Let the foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$ and the hyperbola $\frac{ x ^{2}}{144}-\frac{ y ^{2}}{\alpha}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

medium

(JEE MAIN-2022)

Find the equation of the hyperbola with foci $(0,\,\pm 3)$ and vertices $(0,\,\pm \frac {\sqrt {11}}{2})$.

medium

The eccentricity of the hyperbola $\frac{{{x^2}}}{{16}} – \frac{{{y^2}}}{{25}} = 1$ is

easy

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

hard