Mathematics

Calculus and Analytic Geometry

110 Questions

Calculus and analytic geometry problems involve finding slopes and equations of tangent lines. The focus is on applying derivatives to analyze curves and their geometric properties. These concepts are frequently tested in advanced undergraduate competitive exams.

Tangent line equationsCurve slopesDifferential equationsGeometric curvesNormal to curves

Calculus and Analytic Geometry Questions

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

The focal chord to $y^{2}=16 x$ is tangent to $(x-6)^{2}+y^{2}=2$, then the possible values of the slope of this chord are:

  1. {-1,1}

  2. {-2,2}

  3. {-2,1/2}

  4. (2,-1/2}
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A focal chord of y^2 = 16x passes through (4, 0). The line y - 0 = m(x - 4) is tangent to (x-6)^2 + y^2 = 2. Using the distance from center (6, 0) to the line mx - y - 4m = 0 equal to radius sqrt(2) yields m^2 = 1.

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

the inclination of the tangent at$\theta =\frac { \pi  }{ 3 } on\quad the\quad curve\quad x=a\left( \theta +sin\theta  \right) ,y=a\left( 1+cos\theta  \right) is$

  1. $\frac { \pi }{ 3 }$
  2. $\frac { \pi }{ 6 }$
  3. $\frac { 2\pi }{ 3 }$
  4. $\frac { 5\pi }{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The abscissa of a point on the curve $xy=(a+x)^{2}$, the normal cuts off numerically equal intercepts from the coordinate axes, is

  1. $-\dfrac{a}{\sqrt{2}}$
  2. $\sqrt{2}a$
  3. $\dfrac{a}{\sqrt{2}}$
  4. $-\sqrt{2}a$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
Given,

$xy=(a+x)^2$

differentiating the above equation, we get,

$\Rightarrow y+xy'=2(a+x)$

substituting and solving the above equation, we get,

$\therefore y'=\pm 1$

$y \pm x =2(a+x)$

$\dfrac{(a+x)^2}{x}\pm x=2(a+x)$

$\Rightarrow \pm x=2(a+x)-\dfrac{(a+x)^2}{x}$

$\pm x^2=(2+x)[x-a]$

$\pm x^2=x62-a^2$

$\Rightarrow 2x^2=a^2$

$\therefore x=\pm \dfrac{a}{\sqrt 2}$
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The equation of the curve which is such that the protion of the axis of x cut off between the origin and tangent at any point is proportional to the ordinate of that point is _______________.

  1. $\log x = b y ^ { 2 } + a$
  2. $x = y ( a + b \log y )$
  3. $x = y ( b - a \log y )$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The problem describes a differential equation where the x-intercept of the tangent is proportional to the ordinate y. Solving this leads to the curve x = y(b - a log y).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The length of sub normal to the curve $xy={ a }^{ 2 }$ at (x,y) on it varies at

  1. ${ x }^{ 2 }$
  2. ${ y }^{ 2 }$
  3. ${ x }^{ 3 }$
  4. ${ y }^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$xy=a^2$

$\Rightarrow x\dfrac{dy}{dx}+y=0$

$\therefore \dfrac{dy}{dx}=-\dfrac{y}{x}$

Now,

Sub normal $=y\dfrac{dy}{dx}$

$=y\left ( -\dfrac{y}{x} \right )$

$=-\dfrac{y^2}{x}$

$=-\dfrac{y^2}{\frac{a^2}{y}}$

$=-\dfrac{y^3}{a^2}$

$\therefore SN\propto y^3$
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The normal to a curve at $P(x, y)$ meets the x-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is :

  1. an ellipse

  2. a parabola

  3. a circle

  4. a hyperbola
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Equation of normal $\displaystyle Y-y=-\frac { dy }{ dx } \left( X-x \right) $
$\displaystyle \Rightarrow G=\left( x+y\frac { dy }{ dx } ,0 \right) $
According to question
$\displaystyle \left| x+y\frac { dy }{ dx }  \right| =\left| 2x \right| \Rightarrow y\frac { dy }{ dx } =x$ or $\displaystyle y\frac { dy }{ dx } =-3x$
$\Rightarrow ydy=xdx$ or $ydy=-3xdx$
$\displaystyle \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c$ or $\displaystyle \frac { { y }^{ 2 } }{ 2 } =-\frac { 3{ x }^{ 2 } }{ 2 } +c$
$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }=-2c$ or $3{ x }^{ 2 }+{ y }^{ 2 }=2c$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation ${x}^{2}+9=2{y}^{2}$ is an example of which of the following curves?

  1. hyperbola

  2. circle

  3. ellipse

  4. parabola

  5. line
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, ${x}^{2}+9=2{y}^{2}$ 

$\Rightarrow 2{y}^{2}-{x}^{2}=9$
$\Rightarrow \dfrac { { y }^{ 2 } }{ 9/2 } -\dfrac { { x }^{ 2 } }{ 9 } =1$
It is in the form of $\dfrac { { y }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ { b }^{ 2 } } =1$ which is the equation of hyperbola.
Therefore, the given equation is a equation of hyperbola.

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The angle of intersection at the origin to the curves ${ y }^{ 2 }=4x$ and ${ x }^{ 2 }=4y$ is :

  1. $\pi $
  2. $\dfrac{ \pi }{ 3}$
  3. $\dfrac{ \pi }{ 6 }$
  4. $\dfrac{ \pi }{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given  $y^2=4x$

Differentiating w.r.t. $x$, we get,

$2y\dfrac{dy}{dx}=4$

$\dfrac{dy}{dx}=\dfrac2y.......1$

$x^2=4y$

Differentiating w.r.t. $x$, we get,

$2x=4\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{x}{2}...........2$

So the slope of tangent at $(0,0)$ of $(1)$ is Parallel to $y$ axis 

And the slope of tangent at $(0,0)$ of $(2)$ is Parallel to $x$ axis 

So the angle between them is $\dfrac{\pi}{2}$ 
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the straight lines joining the origin and the points of intersection of the curve $5{x}^{2}+12y-6{y}^{2}+4x-2y+3=0$ and $x+ky-1=0$ are equally inclined to the $x-axis$, then the value of $k$ is equal to:

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

The equation of the pair of lines passing through the origin is obtained by homogenizing the curve equation with the line equation. For the lines to be equally inclined to the x-axis, the coefficient of xy must be zero.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the line $y=x$ cuts the curve ${x}^{3}+{3y}^{3}-30xy+72x-55=0$ in points $A,B$ and $C$ then the value of $\dfrac{4\sqrt{2}}{55}$ $OA.OB.OC$ (where $O$ is the origin ), is ?

  1. $55$
  2. $\dfrac{1}{4\sqrt{2}}$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ x }^{ 3 }+3{ y }^{ 3 }-30xy+72x-55=0$
$y=x$
$\Rightarrow { x }^{ 3 }+3{ x }^{ 3 }-30{ x }^{ 2 }+72x-55=0$
$\Rightarrow 4{ x }^{ 3 }-30{ x }^{ 2 }+72x-55=0$
$\Rightarrow x=1.634,-3.367,2.5$
$\therefore A\left( 1.634,1.634 \right) ;B\left( 3.367,3.367 \right) ;C\left( 2.5,2.5 \right) $
$OA=1.634\sqrt { 2 } ,OB=3.367\sqrt { 2 } ,OC=2.5\sqrt { 2 } $
$=\cfrac { 4\sqrt { 2 }  }{ 55 } \times OA\times OB\times OC=4$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Tangent of the angle at which the curve $y=a^{x}$ and $y=b^{x}(a\neq b>0)$ intersect is given by 

  1. $\dfrac{\log ab}{1+\log ab}$
  2. $\dfrac{\log a/b}{1+\left(\log a\right)\left(\log b\right)}$
  3. $\dfrac{\log ab}{1+\left(\log a\right)\left(\log b\right)}$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection point of y=a^x and y=b^x is (0,1). The slopes of the tangents are ln(a) and ln(b). The tangent of the angle between them is |(ln(a)-ln(b))/(1+ln(a)ln(b))| = |ln(a/b)/(1+ln(a)ln(b))|.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $C$ be a curve which is locus of the point of the intersection of lines $x=2+m$ and $my=4-m$. A circle $s\equiv (x-2)^{2}+(y+1)^{2}=25$ intersector the curve cut at four points $P,Q,R$ and $S$. If $O$ is centre of the curve $C$ the $OP^{2}+OQ^{2}+OR^{2}+OS^{2}$ is

  1. $50$
  2. $100$
  3. $25$
  4. $\dfrac{25}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The locus C is found by eliminating m from x=2+m and my=4-m, yielding (x-2)y = 4-x+2, which simplifies to (x-2)(y+1)=2. This is a rectangular hyperbola centered at (2, -1). For a circle centered at the hyperbola's center, the sum of the squared distances from the center to the intersection points is 4 times the radius squared.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the lines joining the origin to the inter section of the line y = mx+2 and the curve ${ x }^{ 2 }+{ y }^{ 2 }=1$ are at right angles, then

  1. ${ m }^{ 2 }=1$
  2. ${ m }^{ 2 }=3$
  3. ${ m }^{ 2 }=7$
  4. ${ 2m }^{ 2 }=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Homogenizing the equation of the circle x^2 + y^2 = 1 using the line y = mx + 2 (written as (y-mx)/2 = 1) results in x^2 + y^2 = (y-mx)^2/4. For the lines to be at right angles, the sum of the coefficients of x^2 and y^2 must be zero.