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.
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.
since the height and width of the are from the centre is $2\, m$ and $8\, m$ respectively, it is clear that the length of the major axis is $8\, m ,$ while the length of the semi-minor axis is $2 \,m$ The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the $x-$ axis. Hence, the semi-ellipse can be diagrammatically represented as
The equation of the semi-ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 y \geq 0,$ where a is the semimajor axis
Accordingly, $2 a=8 \Rightarrow a=4$ $b=2$
Therefore, the equation of the semi-ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1, y \geq 0$ ....... $(1)$
Let A be a point on the major axis such that $AB =1.5 \,m$
Draw $AC \perp OB$.
$OA =(4-1.5)\, m =2.5 \,m$
The $x-$ coordinate of point $C$ is $2.5$
On substituting the value of $x$ with $2.5$ in equation $(1),$ we obtain
$\frac{(2.5)^{2}}{16}+\frac{y^{2}}{4}=1$
$\Rightarrow \frac{6.25}{16}+\frac{y^{2}}{4}=1$
$\Rightarrow y^{2}=4\left(1-\frac{6.25}{16}\right)$
$\Rightarrow y^{2}=4\left(\frac{9.27}{16}\right)$
$\Rightarrow y^{2}=2.4375$
$\Rightarrow y=1.56 $ (approx.)
$\therefore AC =1.56 \,m$
Thus, the height of the arch at a point $1.5 \,m$ from one end is approximately $1.56 \,m$
Similar Questions
A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{{{x^2}}}{{18}}$ + $\frac{{{y^2}}}{{32}}$ $= 1$ intersects the major and minor axes in points $A$ and $ B$ respectively. If $C$ is the centre of the ellipse then the area of the triangle $ ABC$ is : .............. $\mathrm{sq. \,units}$
A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{{{x^2}}}{{18}}$ + $\frac{{{y^2}}}{{32}}$ $= 1$ intersects the major and minor axes in points $A$ and $ B$ respectively. If $C$ is the centre of the ellipse then the area of the triangle $ ABC$ is : .............. $\mathrm{sq. \,units}$
A ray of light through $(2,1)$ is reflected at a point $P$ on the $y$ - axis and then passes through the point $(5,3)$. If this reflected ray is the directrix of an ellipse with eccentrieity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be :
A ray of light through $(2,1)$ is reflected at a point $P$ on the $y$ - axis and then passes through the point $(5,3)$. If this reflected ray is the directrix of an ellipse with eccentrieity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be :
- [JEE MAIN 2021]
For $0 < \theta < \frac{\pi}{2}$, four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents, the minimum value of $A(\theta)$ is
For $0 < \theta < \frac{\pi}{2}$, four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents, the minimum value of $A(\theta)$ is
- [KVPY 2018]
The minimum area of a triangle formed by any tangent to the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{81}} = 1$ and the coordinate axes is
The minimum area of a triangle formed by any tangent to the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{81}} = 1$ and the coordinate axes is
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $4 x ^{2}+9 y ^{2}=36$
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $4 x ^{2}+9 y ^{2}=36$