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 $\frac{2b^{2}}{a} = \frac{2(5 \cos^{2} 	heta)}{\sqrt{5}} = \frac{10 	imes (2/3)}{\sqrt{5}} = \frac{20}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{3}.$

Column $I$	Column $II$
$(A)$ Circle	$(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$
$(B)$ Parabola	$(q)$ Points $z$ in the complex plane satisfying $\|z+2\|-\|z-2\|= \pm 3$
$(C)$ Ellipse	$(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
$(D)$ Hyperbola	$(s)$ The eccentricity of the conic lies in the interval $1 \leq x < \infty$
	$(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=\|z\|^2+1$

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

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and the parabola $x^2 = 4(y + b)$ are such that the two foci of the ellipse and the end points of the latus rectum of the parabola are the vertices of a square. The eccentricity of the ellipse is

The equation of the common tangent to the circle $x^{2}+y^{2}=2$ and the parabola $y^{2}=8x$ is

$A$ conic passes through the point $(2, 4)$ and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then the foci of the conic are:

Let a tangent to the curve $y^2 = 24x$ meet the curve $xy = 2$ at the points $A$ and $B$. Then the midpoints of such line segments $AB$ lie on a parabola with the

Vedclass Test Series

Exam Paper Generator

Online Exam Module