Answer :

The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive $x$-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form $y^{2}=4 a x$ (as it is opening to the right).

Since the parabola passes through point $A(10,5), 10^{2}=4 a(5)$

$\Rightarrow 100=20 a$

$\Rightarrow a=\dfrac{100}{20}=5$

Therefore, the focus of the parabola is $(a, 0)=(5,0)$, which is the mid-point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

Answer :

The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positivey-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form $x^{2}=4 a y$ (as it is opening upwards).

It can be clearly seen that the parabola passes through point $(\dfrac{5}{2}, 10)$.

$(\dfrac{5}{2})^{2}=4 a(10)$

$\Rightarrow a=\dfrac{25}{4 \times 4 \times 10}=\dfrac{5}{32}$

Therefore, the arch is in the form of a parabola whose equation is

$ x^{2}=\dfrac{5}{8} y $

When $y=2 m, \quad x^{2}=\dfrac{5}{8} \times 2$

$\Rightarrow x^{2}=\dfrac{5}{4}$

$\Rightarrow x=\sqrt{\dfrac{5}{4}} m$

$\therefore AB=2 \times \sqrt{\dfrac{5}{4}} m=2 \times 1.118 m$ (approx.) $=2.23 m$ (approx.)

Hence, when the arch is $2 m$ from the vertex of the parabola, its width is approximately $2.23 m$.

Answer :

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive $y$-axis. This can be diagrammatically represented as

Here, $A B$ and $O C$ are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, $18 m$ from the middle.

Here, $AB=30 m, OC=6 m$, and

$ BC=\dfrac{100}{2}=50 m $

The equation of the parabola is of the form $x^{2}=4 a y$ (as it is opening upwards).

The coordinates of point $A$ are $(50,30$ - 6$)=(50,24)$.

Since $A(50,24)$ is a point on the parabola,

$(50)^{2}=4 a(24)$

$\Rightarrow a=\dfrac{50 \times 50}{4 \times 24}=\dfrac{625}{24}$

$\therefore$ Equation of the parabola, $x^{2}=4 \times \dfrac{625}{24} \times y \quad$ or $6 x^{2}=625 y$

The $x$-coordinate of point $D$ is 18 .

Hence, at $x=18$,

$6(18)^{2}=625 y$

$\Rightarrow y=\dfrac{6 \times 18 \times 18}{625}$

$\Rightarrow y=3.11$ (approx)

$\therefore DE=3.11 m$

$D F=D E+E F=3.11 m+6 m=9.11 m$

Thus, the length of the supporting wire attached to the roadway $18 m$ from the middle is approximately $9.11 m$.

Answer :

Since the height and width of the arc 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 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1, y \geq 0$, where $a$ is the semi-major axis Accordingly, $2 a=8 \Rightarrow a=4$

$b=2$

Therefore, the equation of the semi-ellipse is $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{4}=1, y \geq 0$

Let $A$ be a point on the major axis such that $A B=1.5 m$.

Draw $AC \perp OB$.

$O A=(4 - 1.5) m=2.5 m$

The $x$-coordinate of point $C$ is 2.5 .

On substituting the value of $x$ with 2.5 in equation (1), we obtain

$ \begin{aligned} & \dfrac{(2.5)^{2}}{16}+\dfrac{y^{2}}{4}=1 \\ & \Rightarrow \dfrac{6.25}{16}+\dfrac{y^{2}}{4}=1 \\ & \Rightarrow y^{2}=4(1-\dfrac{6.25}{16}) \\ & \Rightarrow y^{2}=4(\dfrac{9.75}{16}) \\ & \Rightarrow y^{2}=2.4375 \\ & \Rightarrow y=1.56 \quad \text{ (approx.) } \\ & \therefore AC=1.56 m \end{aligned} $

Thus, the height of the arch at a point $1.5 m$ from one end is approximately $1.56 m$.

Answer :

Let $A B$ be the rod making an angle $\theta$ with $O X$ and $P(x, y)$ be the point on it such that $A P=3 cm$.

Then, $P B=A B$ - $A P=(12 - 3) cm=9 cm[A B=12 cm]$

From $P$, draw $PQ \perp OY$ and $PR \perp OX$.

In $\triangle PBQ, \quad \cos \theta=\dfrac{PQ}{PB}=\dfrac{x}{9}$

In $\triangle PRA, \quad \sin \theta=\dfrac{PR}{PA}=\dfrac{y}{3}$

Since, $\sin ^{2} \theta+\cos ^{2} \theta=1$,

$(\dfrac{y}{3})^{2}+(\dfrac{x}{9})^{2}=1$

Or, $\dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}=1$

Thus, the equation of the locus of point $P$ on the rod is $\dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}=1$.

Answer :

The given parabola is $x^{2}=12 y$.

On comparing this equation with $x^{2}=4 a y$, we obtain $4 a=12 \Rightarrow a=3$

$\therefore$ The coordinates of foci are $S(0, a)=S(0,3)$

Let $A B$ be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At $y=3, x^{2}=12(3) \Rightarrow x^{2}=36 \Rightarrow x= \pm 6$

$\therefore$ The coordinates of A are (- 6,3 ), while the coordinates of B are $(6,3)$.

Therefore, the vertices of $\triangle O A B$ are $O(0,0), A(-6,3)$, and $B(6,3)$.

$ \text{ Area of } \begin{aligned} \triangle OAB & =\dfrac{1}{2}|0(3-3)+(-6)(3-0)+6(0-3)| \text{ unit }^{2} \\ & =\dfrac{1}{2}|(-6)(3)+6(-3)| \text{ unit }^{2} \\ & =\dfrac{1}{2}|-18-18| \text{ unit }^{2} \\ & =\dfrac{1}{2}|-36| \text{ unit }^{2} \\ & =\dfrac{1}{2} \times 36 \text{ unit }^{2} \\ & =18 \text{ unit }^{2} \end{aligned} $

Thus, the required area of the triangle is 18 unit $^{2}$.

Answer :

Let $A$ and $B$ be the positions of the two flag posts and $P(x, y)$ be the position of the man. Accordingly, $PA+PB=10$.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is $10 m$, while points $A$ and $B$ are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the $x$-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis

Accordingly, $2 a=10 \Rightarrow a=5$

Distance between the foci $(2 c)=8$

$\Rightarrow c=4$

On using the relation $c=\sqrt{a^{2}-b^{2}}$, we obtain

$4=\sqrt{25-b^{2}}$

$\Rightarrow 16=25-b^{2}$

$\Rightarrow b^{2}=25-16=9$

$\Rightarrow b=3$

Thus, the equation of the path traced by the man is $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1$.

Answer :

Let $OAB$ be the equilateral triangle inscribed in parabola $y^{2}=4 a x$.

Let $A B$ intersect the $x$-axis at point $C$.

Let $OC=k$

From the equation of the given parabola, we have $y^{2}=4 a k \Rightarrow y= \pm 2 \sqrt{a k}$

$\therefore$ The respective coordinates of points $A$ and $B$ are $(k, 2 \sqrt{a k})$, and $(k,-2 \sqrt{a k})$

$AB=CA+CB=2 \sqrt{a k}+2 \sqrt{a k}=4 \sqrt{a k}$

Since $O A B$ is an equilateral triangle, $O A^{2}=A B^{2}$.

$ \begin{aligned} & \therefore k^{2}+(2 \sqrt{a k})^{2}=(4 \sqrt{a k})^{2} \\ & \Rightarrow k^{2}+4 a k=16 a k \\ & \Rightarrow k^{2}=12 a k \\ & \Rightarrow k=12 a \\ & \therefore AB=4 \sqrt{a k}=4 \sqrt{a \times 12 a}=4 \sqrt{12 a^{2}}=8 \sqrt{3} a \end{aligned} $

Thus, the side of the equilateral triangle inscribed in parabola $y^{2}=4$ ax is $8 \sqrt{3} a$.