JEE Main Hyperbola Practice Questions With Solutions

Let the foci of a hyperbola $H$H coincide with the foci of the ellipse $E:\frac{(x-1{)}^2}{100}+\frac{(y-1{)}^2}{75}=1$E : \frac{\left(\right. x - 1 \left.\right)^{2}}{100} + \frac{\left(\right. y - 1 \left.\right)^{2}}{75} = 1 and the eccentricity of the hyperbola $H$H be the reciprocal of the eccentricity of the ellipse $E$E. If the length of the transverse axis of $H$H is $\alpha$\alpha and the length of its conjugate axis is $\beta$\beta, then $3{\alpha }^2+2{\beta }^2$3 \alpha^{2} + 2 \beta^{2} is equal to

Let $H:\frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$H : \frac{- x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 be the hyperbola, whose eccentricity is $\sqrt{3}$\sqrt{3} and the length of the latus rectum is $4\sqrt{3}$4 \sqrt{3}. Suppose the point $(\alpha ,6),\alpha >0$\left(\right. \alpha , 6 \left.\right) , \alpha > 0 lies on $H$H. If $\beta$\beta is the product of the focal distances of the point $(\alpha ,6)$\left(\right. \alpha , 6 \left.\right), then ${\alpha }^2+\beta$\alpha^{2} + \beta is equal to

Consider a hyperbola $\mathrm{H}$H having centre at the origin and foci on the $\mathrm{x}$x-axis. Let ${\mathrm{C}}_1$C_{1} be the circle touching the hyperbola $\mathrm{H}$H and having the centre at the origin. Let ${\mathrm{C}}_2$C_{2} be the circle touching the hyperbola $\mathrm{H}$H at its vertex and having the centre at one of its foci. If areas (in sq units) of $C_1$C_{1} and $C_2$C_{2} are $36\pi$36 \pi and $4\pi$4 \pi, respectively, then the length (in units) of latus rectum of $\mathrm{H}$H is

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1 and the eccentricity of the hyperbola is $\frac{15}{8}$\frac{15}{8} times the eccentricity of the ellipse, then the smaller focal distance of the point $(\sqrt{2},\frac{14}{3}\sqrt{\frac{2}{5}})$\left(\right. \sqrt{2} , \frac{14}{3} \sqrt{\frac{2}{5}} \left.\right) on the hyperbola, is equal to

Let $P$P be a point on the hyperbola $H:\frac{x^2}{9}-\frac{y^2}{4}=1$H : \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1, in the first quadrant such that the area of triangle formed by $P$P and the two foci of $H$H is $2\sqrt{13}$2 \sqrt{13}. Then, the square of the distance of $P$P from the origin is

Let $e_1$e_{1} be the eccentricity of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 and $e_2$e_{2} be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{a}>\mathrm{b}$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 , a > b, which passes through the foci of the hyperbola. If ${\mathrm{e}}_1{\mathrm{e}}_2=1$e_{1} e_{2} = 1, then the length of the chord of the ellipse parallel to the $x$x-axis and passing through $(0,2)$\left(\right. 0 , 2 \left.\right) is :

Let R be a rectangle given by the lines $x=0,x=2,y=0$x = 0 , x = 2 , y = 0 and $y=5$y = 5. Let A$(\alpha ,0)$\left(\right. \alpha , 0 \left.\right) and B$(0,\beta ),\alpha \in [0,2]$\left(\right. 0 , \beta \left.\right) , \alpha \in \left[\right. 0 , 2 \left]\right. and $\beta \in [0,5]$\beta \in \left[\right. 0 , 5 \left]\right., be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :

Let $\mathrm{P}(x_0,y_0)$P \left(\right. x_{0} , y_{0} \left.\right) be the point on the hyperbola $3x^2-4y^2=36$3 x^{2} - 4 y^{2} = 36, which is nearest to the line $3x+2y=1$3 x + 2 y = 1. Then $\sqrt{2}(y_0-x_0)$\sqrt{2} \left(\right. y_{0} - x_{0} \left.\right) is equal to :