NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Conic Section

Question 1: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer:

The parabolic reflector opens towards the right.

So the equation of a parabolic reflector will be,

$y^2=4ax$

Now, since this curve will pass through the point (5,10) if we assume origin at the optical centre,

So

$10^2=4a(5)$

$a=\frac{100}{20}=5$

Hence, the focus of the parabola is,

$(a,0)=(5,0)$.

Alternative Method,

As we know, on any concave curve

$f=\frac{R}{2}$

Hence, Focus

$f=\frac{R}{2}=\frac{10}{2}=5$.

Hence, the focus is 5 cm right to the optical centre.

Question 2: An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer:

Since the Axis of the parabola is vertical, let the equation of the parabola be,

$x^2=4ay$

it can be seen that this curve will pass through the point (5/2, 10) if we assume the origin at the bottom end of the parabolic arch.

So,

$\left(\frac{5}{2}\right)^2=4a(10)$

$a=\frac{25}{160}=\frac{5}{32}$

Hence, the equation of the parabola is

$x^2=4\times\frac{5}{32}\times y$

$x^2=\frac{20}{32} y$

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

Now, when y = 2 the value of x will be

$x=\sqrt{(\frac{5}{8}\times2)}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

Hence, the width of the arch at this height is

$2x=2\times\frac{\sqrt{5}}{2}=\sqrt{5}.$

Question 3: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m Find the length of a supporting wire attached to the roadway 18 m from the middle

Answer:

Given,

The width of the parabolic cable = 100m

The length of the shorter supportive wire attached = 6m

The length of the longer supportive wire attached = 30m

Since the rope opens upwards, the equation will be of the form

$x^2=4ay$

Now, if we consider the origin at the centre of the rope, the equation of the curve will pass through points, (50,30-6)=(50,24)

$24^2=4a50$

$a=\frac{625}{24}$

Hence, the equation of the parabola is

$x^2=4\times \frac{625}{24}\times y$

$x^2= \frac{625}{6}\times y$

Now, at a point 18 m right from the centre of the rope, the x coordinate of that point will be 18, so by the equation, the y coordinate will be

$y=\frac{x^2}{4a}=\frac{18^2}{4\times \frac{625}{6}}\approx 3.11m$

Hence, the length of the supporting wire attached to the roadway from the middle is 3.11+6=9.11m.

Question 4: An arch is in the form of a semi-ellipse. 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.

Answer:

The equation of the semi-ellipse will be of the form

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\:,y>0$

Now, according to the question,

the length of major axis = 2a = 8 $\Rightarrow a=4$

The length of the semimajor axis =2$\Rightarrow b=2$

Hence, the equation will be,

$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1\:,y>0$

$\frac{x^2}{16}+\frac{y^2}{4}=1\:,y>0$

Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5

So, the height at this point is

$\frac{(2.5)^2}{16}+\frac{y^2}{4}=1\Rightarrow y=\sqrt{4(1-\frac{2.5^2}{16})}$

$y\approx 1.56m$

Hence, the height of the required point is 1.56 m.

Question 5: A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Answer:

Let $\theta$ be the angle that rod makes with the ground,

Now, at a point 3 cm from the end,

$\cos\theta=\frac{x}{9}$

At the point touching the ground

$\sin\theta=\frac{y}{3}$

Now, as we know the trigonometric identity,

$\sin^2\theta+\cos^2\theta=1$

$\left (\frac{x}{9} \right )^2+\left ( \frac{y}{3} \right )^2=1$

$\frac{x^2}{81}+\frac{y^2}{9}=1$

Hence, the equation is,

$\frac{x^2}{81}+\frac{y^2}{9}=1$

Question 6: Find the area of the triangle formed by the lines joining the vertex of the parabola $x ^2 = 12y$ to the ends of its latus rectum.

Answer:

Given the parabola,

$x^2=12y$

Comparing this equation with $x^2=4ay$, we get

$a=3$

Now, as we know, the coordinates of the ends of the latus rectum are

$(2a,a)\:and\:(-2a,a)$

So, the coordinates of the latus rectum are,

$(2a,a)\:and\:(-2a,a)=(6,3)\:and\:(-6,3)$

Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3)

Width of the triangle = 2*6=12

Height of the triangle = 3

So the area =

$\frac{1}{2}\times base\times height=\frac{1}{2}\times12\times3=18$

Hence, the required area is 18 units square.

Question 7: A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Answer:

As we know, if a point moves in a plane in such a way that its distance from two points remains constant, then the path is an ellipse.

Now, according to the question,

The distance between the point where the sum of the distances from a point is constant = 10

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

Now, the distance between the foci=8

$\Rightarrow 2c=8\Rightarrow c=4$

Now, as we know the relation,

$c^2=a^2-b^2$

$b^2=a^2-c^2$

$b=\sqrt{a^2-c^2}=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3$

Hence, the equation of the ellipse is,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$

Hence, the path of the man will be

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$

Question 8: An equilateral triangle is inscribed in the parabola $y^2 = 4 ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer:

Given an equilateral triangle inscribed in a parabola with the equation.$y^2 = 4 ax$

One coordinate of the triangle is A(0,0).

Now, let the other two coordinates of the triangle be

$B(x,\sqrt{4ax})$ and $C(x,-\sqrt{4ax})$

Now, since the triangle is equilateral,

$BC=AB=CA$

$2\sqrt{4ax}=\sqrt{(x-0)^2+(\sqrt{4ax}-0)^2}$

$x^2=12ax$

$x=12a$

The coordinates of the points of the equilateral triangle are,

$(0,0),(12,\sqrt{4a\times 12a}),(12,-\sqrt{4a\times 12a})=(0,0),(12,4\sqrt{3}a)\:and\:(12,-4\sqrt{3}a)$

So, the side of the triangle is

$2\sqrt{4ax}=2\times4\sqrt{3}a=8\sqrt{3}a$