Let the hyperbola H : frac{x^{2}}{a^{2}}-frac | vedclass

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

Question

(B) The hyperbola is $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$.For the parabola,the focus is $(ae, 0)$

Vedclass · Accepted Answer

$(3\sqrt{3}, -6\sqrt{2})$

Vedclass · Answer

(B) The hyperbola is $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. The foci are $S(ae, 0)$ and $S'(-ae, 0)$.For the parabola,the focus is $(ae, 0)$ and the directrix is $x = -ae$.The distance between the focus and the directrix is $2ae$. Since the distance between the focus and directrix of a parabola is $2p$ (where $4p$ is the latus rectum),we have $2p = 2ae$,so $p = ae$.The length of the latus rectum of the parabola is $4p = 4ae$.The length of the latus rectum of $H$ is $\frac{2b^{2}}{a}$.Given $4ae = e 	imes \frac{2b^{2}}{a}$,we get $4a = \frac{2b^{2}}{a}$,so $b^{2} = 2a^{2}$.Since $(2\sqrt{2}, -2\sqrt{2})$ lies on $H$,$\frac{8}{a^{2}} - \frac{8}{b^{2}} = 1$. Substituting $b^{2} = 2a^{2}$,we get $\frac{8}{a^{2}} - \frac{4}{a^{2}} = 1$,so $a^{2} = 4$ and $b^{2} = 8$.Then $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + 2 = 3$,so $e = \sqrt{3}$.The focus of the parabola is $(ae, 0) = (2\sqrt{3}, 0)$ and the directrix is $x = -2\sqrt{3}$.The equation of the parabola is $(y-0)^{2} = 4(ae)(x - (-ae)) = 4(2\sqrt{3})(x + 2\sqrt{3})$ is incorrect based on standard form. Using $y^{2} = 4p(x - h)$,where $p = ae = 2\sqrt{3}$ and vertex is at origin relative to focus/directrix,the parabola is $y^{2} = 4(2\sqrt{3})x = 8\sqrt{3}x$.Testing points: For $(3\sqrt{3}, -6\sqrt{2})$,$y^{2} = (-6\sqrt{2})^{2} = 72$ and $8\sqrt{3}x = 8\sqrt{3}(3\sqrt{3}) = 72$. Thus,$(3\sqrt{3}, -6\sqrt{2})$ lies on the parabola.

Similar Questions

The tangent at $P$ to a parabola $y^2 = 4ax$ meets the directrix at $U$ and the latus rectum at $V$. Then $\triangle SUV$ (where $S$ is the focus) :

If $p$ and $q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively,then the area of the square (in sq. units) formed by the points of intersection of the ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ and the pair of lines $x^2-y^2=0$ is

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

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be an ellipse,whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module