Let the foci of a hyperbola H coincide with t | vedclass

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

Question

(B) For the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$,we have $a^2=100$ and $b^2=75$.The eccentricity $e_1 = \sqrt{1-\frac{b^2}{a^2}} = \s

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(B) For the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$,we have $a^2=100$ and $b^2=75$.The eccentricity $e_1 = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{75}{100}} = \sqrt{\frac{25}{100}} = \frac{5}{10} = \frac{1}{2}$.The foci of the ellipse are $(h \pm ae_1, k) = (1 \pm 10 	imes \frac{1}{2}, 1) = (1 \pm 5, 1)$,which are $F_1(6, 1)$ and $F_2(-4, 1)$.The distance between the foci is $2ae_1 = 10$.For the hyperbola $H$,the eccentricity $e_2 = \frac{1}{e_1} = 2$.The distance between the foci of the hyperbola is $2ae_2 = 10$,so $2a(2) = 10$,which gives $a = \frac{5}{2}$.The length of the transverse axis is $\alpha = 2a = 5$.For a hyperbola,$b^2 = a^2(e_2^2-1) = a^2(2^2-1) = 3a^2$.Thus,$b = a\sqrt{3} = \frac{5\sqrt{3}}{2}$.The length of the conjugate axis is $\beta = 2b = 5\sqrt{3}$.Therefore,$3\alpha^2 + 2\beta^2 = 3(5)^2 + 2(5\sqrt{3})^2 = 3(25) + 2(75) = 75 + 150 = 225$.

Similar Questions

For what value of $m$ is the line $y = mx + 6$ a tangent to the hyperbola $\frac{x^2}{100} - \frac{y^2}{49} = 1$?

If $x = 9$ is the chord of contact of the hyperbola $x^2 - y^2 = 9$,then the equation of the corresponding pair of tangents is

If the circle $x^{2}+y^{2}=a^{2}$ intersects the hyperbola $xy=c^{2}$ in four points $P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3})$ and $S(x_{4}, y_{4})$,then

If the tangent drawn at a point $P(t)$ on the hyperbola $x^2-y^2=c^2$ cuts the $X$-axis at $T$ and the normal drawn at the same point $P$ cuts the $Y$-axis at $N$,then the equation of the locus of the midpoint of $TN$ is

If the eccentricity of a hyperbola $\frac{x^2}{9} - \frac{y^2}{b^2} = 1,$ which passes through $(K, 2),$ is $\frac{\sqrt{13}}{3},$ then the value of $K^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module