The tangent at an extremity (in the first qua | vedclass

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

Question

(A) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$.Here,$a^2 = 4$ and $b^2 = 5$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 +

Vedclass · Accepted Answer

$-\frac{20}{9}$

Vedclass · Answer

(A) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$.Here,$a^2 = 4$ and $b^2 = 5$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.The extremity of the latus rectum in the first quadrant is $L = (ae, \frac{b^2}{a}) = (2 	imes \frac{3}{2}, \frac{5}{2}) = (3, \frac{5}{2})$.The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.Substituting $(x_1, y_1) = (3, \frac{5}{2})$,we get $\frac{3x}{4} - \frac{y(5/2)}{5} = 1$,which simplifies to $\frac{3x}{4} - \frac{y}{2} = 1$.This can be written as $\frac{x}{4/3} + \frac{y}{-2} = 1$.The $x$-intercept is $OA = \frac{4}{3}$ and the $y$-intercept is $OB = -2$.Therefore,$(OA)^2 - (OB)^2 = (\frac{4}{3})^2 - (-2)^2 = \frac{16}{9} - 4 = \frac{16 - 36}{9} = -\frac{20}{9}$.

The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. Then $(OA)^2 - (OB)^2$,where $O$ is the origin,equals

Similar Questions

The point of intersection of two tangents drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$ lies on the circle $x^2 + y^2 = 5$. If these tangents are perpendicular to each other,then $a =$

The locus of the point of intersection of tangents at the ends of a normal chord of the hyperbola $x^2 - y^2 = a^2$ is

If for a hyperbola the ratio of the length of the conjugate axis to the length of the transverse axis is $3:2$,then the ratio of the distance between the foci to the distance between the two directrices is

The eccentricity of the hyperbola $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

If the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{5}{4}$ and $2x+3y-6=0$ is a focal chord of the hyperbola,then the length of the transverse axis is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module