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From the perfect curve of a football’s path to the elliptical orbits of planets, conic sections are all around us. These curves- circle, ellipse, parabola, and hyperbola are formed by slicing a cone at different angles. In mathematics, they will help you to model motion, design structures and even focus light and sound! In this exercise, you will apply all the key concepts of conic sections, like equations of circles, ellipses, parabolas, hyperbolas, etc. The questions are a blend of geometry and algebra that will test your overall understanding of the chapter.
The NCERT is a key guide that offers step-by-step calculations and clear explanations to help you tackle a variety of problems. These NCERT solutions will help you improve your speed and accuracy and will prepare you for upcoming exams.
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.
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
$2x=2\times\frac{\sqrt{5}}{2}=\sqrt{5}.$
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.
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.
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,
$\frac{x^2}{81}+\frac{y^2}{9}=1$
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.
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
$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$
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
$2\sqrt{4ax}=2\times4\sqrt{3}a=8\sqrt{3}a$
Also read
1. Circle
A circle is the set of all points in a plane that are equidistant from a fixed point called the center.
$$
(x-h)^2+(y-k)^2=r^2
$$
where $(h, k)$ is the center and $r$ is the radius.
2. Parabola
The collection of all points that are equally spaced between a given line and a fixed point (focus) is called a parabola.
Standard equation (horizontal axis) is given by
$$
y^2=4 a x
$$
3. Ellipse
An ellipse is the set of all points in a plane such that the sum of distances from two fixed points (called foci) is constant.
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
4. Hyperbola
The collection of all points in a plane where the distance difference between two fixed points (foci) is constant is called a hyperbola.
$$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$$
5. Latus Rectum
A line segment that runs through the focus and is perpendicular to a conic's axis of symmetry is called the latus rectum. It helps in defining the width of a conic near the focus.
Also read
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Equation of circle => $(x-1)^2 + (y-2)^2 = 5^2$
$x^2 -2x + 1 + y^2 + 4y + 4 = 25$
$x^2 -2x + y^2 + 4y = 20$
Equation of the parabola => $y^2 = 4(2)x$
$y^2 = 8x$
Given equation of the parabola => $y^2 = 12x$
Compare with $y^2 = 4ax$ => a = 3
Coordinate of focus is (3,0).
The given equation involves $y^2$, so the axis of symmetry is along the x-axis. The axis of the parabola is the x-axis.
Given equation of the parabola => $y^2 = 12x$
Compare with $y^2 = 4ax$ => a = 3
The equation of the directrix of the parabola is x = – 3
Given equation of the parabola => $y^2 = 12x$
Compare with $y^2 = 4ax$ => a = 3
Length of the latus rectum is 4a = 4 × 3 = 12.
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