Tag: fundamental concepts - geometry

Quizzes (4) Questions (19)

Questions Related to fundamental concepts - geometry

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The straight line AB is divided at C so that $\bar{AC} = 3\bar{CB}$. Circles are described on AC and CB as diameters and a common tangent meets AB produced at D. Then $\bar{BD}$ equals.

  1. the diameter of the smaller circle

  2. the radius of the smaller circle

  3. the radius of the larger circle
  4. $\bar{CB} \sqrt{3}$

  5. the difference of the two radii
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $ x=\overline{BD} $ and let $ r$ be the radius of the small circle. 


Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. 

By similar triangles, 

$ \dfrac{x+r}{r}=\dfrac{x+5r}{3r} \implies x=r$.

$ \overline{BD} $ equals the radius of the smaller circle.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

All chords of the curve $x^{2}+y^{2}-10x-4y+4=0$  which make a right angle at $(8,2)$ pass through

  1. $(2,5)$
  2. $(-2,-5)$
  3. $(-5,-2)$
  4. $(5,2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a circle x^2 + y^2 - 10x - 4y + 4 = 0, the center is (5, 2). Chords subtending a right angle at a point P(8, 2) pass through a fixed point. By the property of circles, if a chord subtends a right angle at P, the locus of the intersection of tangents at the chord's endpoints is related to the circle's geometry. The fixed point is the center (5, 2).

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

Let $m$ be the slope of tangent to the curve $e^{2y}=1+x^{2}$ then set of all values  of $m$ is :

  1. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$
  2. $\left[-\infty, -\dfrac{1}{2}\right]\cup \left[\dfrac{1}{2}, 0\right]$
  3. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]-\left\{0\right\}$
  4. $[-2,2]-\left\{0\right\}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m$ be the slope of tangent.

Given equation of curve is

${{e}^{2y}}=1+{{x}^{2}}$


Taking log both side and we get,

$ \log {{e}^{2y}}=\log \left( 1+{{x}^{2}} \right) $

$ 2y=\log \left( 1+{{x}^{2}} \right) $


On differentiating and we get,

$ 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}2x $

$ \Rightarrow \dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}} $

$m=\dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}}$


On put $x=\left( -1,1 \right)$

So,

$ m=\dfrac{1}{1+1}=\dfrac{1}{2} $

$ m=\dfrac{-1}{1+1}=\dfrac{-1}{2} $

Hence, the value of $m$ is $\left[-\dfrac{1}{2},\dfrac{1}{2} \right]$


Hence, this is the answer.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The radius of the locus by the point represented by $z$, when $arg\dfrac {z-1}{z+1} =\dfrac {\pi}{4}$, is

  1. $\sqrt {2}$
  2. $\sqrt {2}\pi$
  3. $\dfrac {\pi}{\sqrt {2}}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Arg$\left[\cfrac{z-1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\times \cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\right] \times$Arg$\left[\cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$ Arg${\left[z-1\right]}-$Arg${\left[z+1\right]}=\dfrac{\pi}{4}$

Let $z=x+iy$ 

$\therefore$Arg$\left[\left(x-1\right)+iy\right]-$Arg$\left[\left(x+1\right)+iy\right]=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}\left(\dfrac{y}{x-1}\right)-{\tan}^{-1}\left(\dfrac{y}{x+1}\right)=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}{\left(\dfrac{\dfrac{y}{x-1}-\dfrac{y}{x+1}}{1+\dfrac{{y}^{2}}{{x}^{2}-1}}\right)}=\dfrac{\pi}{4}$

$\Rightarrow \dfrac{2y}{{x}^{2}-1+{y}^{2}}=1$

$\Rightarrow {x}^{2}-1+{y}^{2}=2y$

$\Rightarrow {x}^{2}-2y+{y}^{2}-1=0$

This represents a circle with center at $\left(0,1\right)$ and radius  $=\sqrt{0+1-\left(-1\right)}=\sqrt{2}$
Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

A curve which begins and ends at the same point is called a:

  1. closed curve

  2. open curve

  3. normal curve

  4. definite curve
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A close curve is made up of a closed boundary. It initialised by a fixed point and end with the same point.

So, a curve which begins and ends at the same point is called a closed curve.
Hence, the answer is a closed curve.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

A point is moving along the curve ${ y }^{ 3 }=27x$. Find the interval of valued of $x$ in which the ordinate changes faster then abscissa is:

  1. $x\in \left( -1,1 \right)$
  2. $x\in \left( -1,-1 \right) -\left\{ 0 \right\}$
  3. $x\in \left[ -1,1 \right] -\left\{ 0 \right\}$
  4. $x\in \left( -1,0 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$y^3=27x$

Abcissa changes at slower rate than ordinate.
$\dfrac{dx}{dt}<\dfrac{dy}{dt}.................(1)$
$y^3=27x$
$3y^2\dfrac{dy}{dt}=27\dfrac{dx}{dt}.............(2)$
Putting $\dfrac{dx}{dt}$ in eq $(1)$
 $\dfrac{3y^2}{27}<\dfrac{dy}{dy}$
 $\dfrac{dy}{dt}\left(\dfrac{3y^2}{27}<1\right)<0$
By eq$(2)$ wecan say that $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ will be '+ve' or '-ve'.
So, $\dfrac{3y^2}{27}-1<0\Rightarrow{-3}<y<3$ and $-1<x<1$.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The curves $y = 2{\left( {x - a} \right)^2}andy = {e^{2x}}$ touches each other, then'a' is less than- 

  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the curves to touch, they must share a common tangent at a point. Setting y = 2(x-a)^2 and y = e^(2x) equal and their derivatives equal: 2(x-a)^2 = e^(2x) and 4(x-a) = 2e^(2x). Solving these leads to the condition for 'a'.