For some theta in left(0, frac{pi}{2}right), | vedclass

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

Question

(B) Given the hyperbola equation: $x^{2} - y^{2} \sec^{2} 	heta = 10 \Rightarrow \frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} 	heta} = 1.$For a hyper

Vedclass · Accepted Answer

$\frac{4 \sqrt{5}}{3}$

Vedclass · Answer

(B) Given the hyperbola equation: $x^{2} - y^{2} \sec^{2} 	heta = 10 \Rightarrow \frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} 	heta} = 1.$For a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ the eccentricity $e_{H} = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$Thus,$e_{H} = \sqrt{1 + \frac{10 \cos^{2} 	heta}{10}} = \sqrt{1 + \cos^{2} 	heta}.$Given the ellipse equation: $x^{2} \sec^{2} 	heta + y^{2} = 5 \Rightarrow \frac{x^{2}}{5 \cos^{2} 	heta} + \frac{y^{2}}{5} = 1.$For an ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$ (where $a > b$),$e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}}}.$Here $a^{2} = 5$ and $b^{2} = 5 \cos^{2} 	heta,$ so $e_{E} = \sqrt{1 - \frac{5 \cos^{2} 	heta}{5}} = \sqrt{1 - \cos^{2} 	heta} = \sin 	heta.$Given $e_{H} = \sqrt{5} e_{E},$ we have $\sqrt{1 + \cos^{2} 	heta} = \sqrt{5} \sin 	heta.$Squaring both sides: $1 + \cos^{2} 	heta = 5 \sin^{2} 	heta = 5(1 - \cos^{2} 	heta).$$1 + \cos^{2} 	heta = 5 - 5 \cos^{2} 	heta$ $\Rightarrow 6 \cos^{2} 	heta = 4$ $\Rightarrow \cos^{2} 	heta = \frac{2}{3}.$Then $\sin^{2} 	heta = 1 - \frac{2}{3} = \frac{1}{3}.$The length of the latus rectum of the ellipse is $\frac{2b^{2}}{a} = \frac{2(5 \cos^{2} 	heta)}{\sqrt{5}} = \frac{10 	imes (2/3)}{\sqrt{5}} = \frac{20}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{3}.$

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

For some $\theta \in \left(0, \frac{\pi}{2}\right),$ if the eccentricity of the hyperbola $x^{2} - y^{2} \sec^{2} \theta = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse $x^{2} \sec^{2} \theta + y^{2} = 5,$ then the length of the latus rectum of the ellipse is

Similar Questions

The locus of the midpoints of the chords of the hyperbola $x^{2}-y^{2}=4$,which touch the parabola $y^{2}=8x$,is:

$PQ$ is a double ordinate of the parabola $y^2 = 4ax$. What is the locus of the point of intersection of the normals at $P$ and $Q$?

Suppose $A, B, C$ and $D$ are the four intersection points of the curves $\frac{x^2}{18}+\frac{y^2}{8}=1$ and $x^2-y^2=5$ in $I, II, III$ and $IV$ quadrants respectively. If $\theta_1, \theta_2, \theta_3$ and $\theta_4$ respectively are the angles between the curves at $A, B, C$ and $D$, then

If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide,then $b^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module