Let $a$ and $b$ respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9e^2 - 18e + 5 = 0$. If $S(5, 0)$ is a focus and $5x = 9$ is the corresponding directrix of this hyperbola,then $a^2 - b^2$ is equal to
- A$-7$
- B$-5$
- C$5$
- D$7$
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If $x-y=1$ is a tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$,then the point of contact is
MediumKCET 2012
View SolutionLet $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a > b > 0$,be a hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4\sqrt{3}$.
List-$I$ List-$II$ $P$. The length of the conjugate axis of $H$ is $1$. $8$ $Q$. The eccentricity of $H$ is $2$. $\frac{4}{\sqrt{3}}$ $R$. The distance between the foci of $H$ is $3$. $\frac{2}{\sqrt{3}}$ $S$. The length of the latus rectum of $H$ is $4$. $4$
The correct option is:
| List-$I$ | List-$II$ |
| $P$. The length of the conjugate axis of $H$ is | $1$. $8$ |
| $Q$. The eccentricity of $H$ is | $2$. $\frac{4}{\sqrt{3}}$ |
| $R$. The distance between the foci of $H$ is | $3$. $\frac{2}{\sqrt{3}}$ |
| $S$. The length of the latus rectum of $H$ is | $4$. $4$ |
The correct option is:
The equation of the normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ that is perpendicular to the line $2x + y = 1$ 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 $\beta=$
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