With the origin as a focus and x = 4 as the c | vedclass

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

Question

(B) Let the focus be $F(0, 0)$ and the directrix be $x = 4$. Let the end of the minor axis be $B(h, k)$.By the definition of an ellipse, the distance

Vedclass · Accepted Answer

a parabola

Vedclass · Answer

(B) Let the focus be $F(0, 0)$ and the directrix be $x = 4$. Let the end of the minor axis be $B(h, k)$.By the definition of an ellipse, the distance from the focus to the point $B$ is $e$ times the distance from $B$ to the directrix.$BF = e \cdot BM$$BF = \sqrt{h^2 + k^2}$ and $BM = |4 - h|$.So, $\sqrt{h^2 + k^2} = e(4 - h)$.Also, for an ellipse, the distance from the center to the focus is $ae$ and the distance from the center to the directrix is $a/e$. Since the focus is at $(0,0)$ and the directrix is $x=4$, the distance between them is $a/e - ae = 4$.For the end of the minor axis, the distance from the center to the focus is $ae$, so $h = -ae$ and $k = b$.From $a/e - ae = 4$, we have $a(1 - e^2) = 4e$, which implies $b^2/a = 4e$, so $b^2 = 4ae$.Substituting $h = -ae$ and $k = b$, we get $k^2 = 4(-h) = -4h$.Replacing $(h, k)$ with $(x, y)$, the locus is $y^2 = -4x$, which is a parabola.

With the origin as a focus and $x = 4$ as the corresponding directrix, a family of ellipses is drawn. Then the locus of an end of the minor axis is

Similar Questions

The foci of the ellipse $25(x + 1)^2 + 9(y + 2)^2 = 225$ are at

Let $F_1$ and $F_2$ be the foci of an ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$. $A$ ray from $F_1$ strikes the elliptical mirror at point $P$ and is reflected. What is the equation of the angle bisector of the angle between the incident ray and the reflected ray?

Find the area of the quadrilateral formed by the tangents at the endpoints of the latus rectum of the ellipse $x^2 + 2y^2 = 2$.

If an ellipse with its axes as coordinate axes,$2a$ and $2b$ as the lengths of its major and minor axes respectively,passes through the points $(2,2)$ and $(3,1)$,then $3a^2+5b^2=$

The product of the lengths of perpendiculars from the foci on any tangent to the ellipse $3x^2 + 5y^2 = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module