Find the equation of the ellipse, whose length of the major axis is $20$ and foci are $(0,\,\pm 5)$
since the foci are on $y-$ axis, the major axis is along the $y-$ axis. So, equation of the cllipse is of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$
Given that
$a=$ semi-major axis $=\frac{20}{2}=10$
and the relation $c^{2}=a^{2}-b^{2}$ gives
$5^{2}=10^{2}-b^{2} $ i.e., $b^{2}=75$
Therefore, the equation of the ellipse is
$\frac{x^{2}}{75}+\frac{y^{2}}{100}=1$
The the circle passing through the foci of the $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$ and having centre at $(0,3) $ is
If $P_1$ and $P_2$ are two points on the ellipse $\frac{{{x^2}}}{4} + {y^2} = 1$ at which the tangents are parallel to the chord joining the points $(0, 1)$ and $(2, 0)$, then the distance between $P_1$ and $P_2$ is
The line $x\cos \alpha + y\sin \alpha = p$ will be a tangent to the conic $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if
An arch is in the form of a semi-cllipse. It is $8 \,m$ wide and $2 \,m$ high at the centre. Find the height of the arch at a point $1.5\, m$ from one end.
Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6,\,0),$ foci $(\pm 4,\,0)$