A hyperbola passes through the point P(sqrt{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given foci are $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$. Also,$b^2 = a^2(e

Vedclass · Accepted Answer

$(2\sqrt{2}, 3\sqrt{3})$

Vedclass · Answer

(C) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given foci are $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$. Also,$b^2 = a^2(e^2 - 1) = a^2e^2 - a^2 = 4 - a^2$. Since the hyperbola passes through $P(\sqrt{2}, \sqrt{3})$,we have $\frac{2}{a^2} - \frac{3}{4 - a^2} = 1$. Let $u = a^2$. Then $\frac{2}{u} - \frac{3}{4 - u} = 1 \implies 2(4 - u) - 3u = u(4 - u) \implies 8 - 2u - 3u = 4u - u^2 \implies u^2 - 9u + 8 = 0$. Solving for $u$,$(u - 8)(u - 1) = 0$. Since $a^2 Then $b^2 = 4 - 1 = 3$. The equation is $x^2 - \frac{y^2}{3} = 1$. The tangent at $P(\sqrt{2}, \sqrt{3})$ is $x(\sqrt{2}) - \frac{y(\sqrt{3})}{3} = 1$,which simplifies to $\sqrt{2}x - \frac{y}{\sqrt{3}} = 1$. Checking the options: For option $D$,$x = 3\sqrt{2}$ and $y = 2\sqrt{3}$,we get $\sqrt{2}(3\sqrt{2}) - \frac{2\sqrt{3}}{\sqrt{3}} = 6 - 2 = 4 
eq 1$. Re-evaluating: The tangent is $\sqrt{2}x - \frac{y}{\sqrt{3}} = 1$. Substituting $x = 2\sqrt{2}, y = 3\sqrt{3}$ gives $\sqrt{2}(2\sqrt{2}) - \frac{3\sqrt{3}}{\sqrt{3}} = 4 - 3 = 1$. Thus,the point $(2\sqrt{2}, 3\sqrt{3})$ lies on the tangent.

$A$ hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the point that lies on the tangent drawn to this hyperbola at $P$ is

Similar Questions

Let $A(-1, 0)$ and $B(2, 0)$ be two points. $A$ point $M$ moves in the plane in such a way that $\angle MBA = 2 \angle MAB$. Then, the point $M$ moves along

$A$ hyperbola having its centre at the origin passes through the point $(5, 2)$ and has a transverse axis of length $8$ along the $X$-axis. What is the eccentricity of its conjugate hyperbola?

The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ form with any tangent to the hyperbola a triangle whose area is $a^2 \tan \lambda$ in magnitude. Then its eccentricity $e$ is:

If the tangent drawn at the point $P(3 \sqrt{2}, 4)$ on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ meets its directrix at $Q(\alpha, \beta)$ in the fourth quadrant,then $\beta=$

For all real values of $m$,the straight line $y = mx + \sqrt{9m^2 - 4}$ is a tangent to the curve:

Vedclass Test Series

Exam Paper Generator

Online Exam Module