Embibe Experts Solutions for Chapter: Hyperbola, Exercise 1: Exercise 1

A point $P$ is taken on the right half of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having its foci as $S_1$ and $S_2.$ If the internal angle bisector of the angle $\angle S_{1}P{S}_2$ cuts the $x$ -axis at the point $Q(\alpha ,0)$, then $\alpha \in$

If $P\mathit{(}\alpha \mathit{,}\mathit{ }\beta \mathit{)}\mathit{,}$ the point of intersection of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{a^2\left(1-e^2\right)}=1$ and the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{a^2\left(E^2-1\right)}=\frac{1}{4}$ is equidistant from the foci of the two curves (all lying on the right of $y$-axis), then

A hyperbola having transverse axis of length $\frac{1}{2}$ unit is confocal with the ellipse $3x^2+4y^2=12$. Then

The equation of a hyperbola with co-ordinate axes as principal axes, and the distances of one of its vertices from the foci are $3$ and $1$, can be

If $2x-y+1=0$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{16}=1$ then which of the following CANNOT be sides of a right angled triangle?

The normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the axes in $M$ and $N$, and lines $MP$ and $NP$ are drawn at right angle to the axes then locus of $P$ is a hyperbola with eccentricity $e\text{'}$, if eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $e$ then

If the axis of a varying central hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be fixed in magnitude and position, then locus of point of contact of a tangent drawn to it from a fixed point on $x$-axis is 

A square $ABCD$ has all its vertices on the curve $x^2{y}^2=1.$ The midpoints of its sides also lie on the same curve. Then, the square of area of $ABCD$ is