Find the equation for the ellipse that satisfies the given conditions: $b=3,\,\, c=4,$ centre at the origin; foci on the $x$ axis.
It is given that $b=3,\,\, c=4,$ centre at the origin; foci on the $x$ axis.
since the foci are on the $x-$ axis, the major axis is along the $x-$ axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ where a is the semimajor axis.
Accordingly, $b=3, \,\,c=4$
It is known that $a^{2}=b^{2}+c^{2}$
$\therefore a^{2}=3^{2}+4^{2}=9+16=25$
$\Rightarrow a=5$
Thus, the equation of the ellipse is $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ or $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is
Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1<0, y_2<0$, be the end points of the latus rectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latus rectum $P Q$ are
$(A)$ $x^2+2 \sqrt{3} y=3+\sqrt{3}$
$(B)$ $x^2-2 \sqrt{3} y=3+\sqrt{3}$
$(C)$ $x^2+2 \sqrt{3} y=3-\sqrt{3}$
$(D)$ $x^2-2 \sqrt{3} y=3-\sqrt{3}$
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The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $P Q$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points
Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 5,\,0),$ foci $(\pm 4,\,0)$