Thank you for reporting, we will resolve it shortly
Q.
An arch is in the form of semi-ellipse. It is $8 \,m$ wide and $2\, m$ high at the centre. Then, the height of the arch at a point $1.5\, m$ from one end is
Conic Sections
Solution:
Clearly, equation of ellipse takes the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \,\,\,\,\,\,\, ...(i)$
Here, it is given $2 a =8$ and $b =2 \Rightarrow a =4, b =2$
Put the values of a and b in eq. (i), we get
$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$
Given, $AP =1.5 m \Rightarrow OP = OA - AP =4-1.5$
$\Rightarrow OP =2.5 M$
Let $PQ = k$
$\therefore $ Coordinate $Q =(2.5, k )$ will satisfy the equation of ellipse.
i.e.,$\frac{(2.5)^{2}}{16}+\frac{ k ^{2}}{4}=1 \Rightarrow \frac{6.25}{16}+\frac{ k ^{2}}{4}=1$
$ \Rightarrow \frac{ k ^{2}}{4}=\frac{1}{1}-\frac{6.25}{16}=\frac{16-6.25}{16}$
$\frac{ k ^{2}}{4}=\frac{9.75}{16} \Rightarrow k ^{2}=\frac{9.75}{4}$
$\Rightarrow k ^{2}=2.4375 \Rightarrow k =1.56\, m ($ approx. $)$
