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 maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The new equation of the curve $4(x-2y+1)^{2}+9(2x+y+2)^{2}=25$ if the lines $2x+y+2=0$ and $x-2y+1=0$ are taken as the new $x$ and $y$ axes respectively is

  1. $4X^{2}+9Y^{2}=5$
  2. $4X^{2}+9Y^{2}=25$
  3. $4X^{2}+9Y^{2}=7$
  4. $4X^{2}-9Y^{2}=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By substituting the new axes X = 2x + y + 2 and Y = x - 2y + 1 into the equation, the expression simplifies directly to 4Y^2 + 9X^2 = 25. The question asks for the equation in terms of X and Y.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If $F _{1}=\left ( 3, 0 \right )$, $F _{2}=\left ( -3, 0 \right )$ and $P$ is any point on the curve $16x^{2}+25y^{2}=400$, then $PF _{1}+PF _{2}$ equals to:

  1. $8$
  2. $6$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the ellipse can be written as $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$

Here $a^{2}=25$, $b^{2}=16$

But $b^{2}=a^{2}\left ( 1-e^{2} \right )$

$\Rightarrow $   $16=25\left ( 1-e^{2} \right )$   $\Rightarrow $   $e=\dfrac35$

So that foci of the ellipse are $\left ( \pm ae, 0 \right )$ i.e. $\left ( \pm 3, 0 \right )$ or $F _{1}$ and $F _{2}.$

By definition of the ellipse, since $P$ is any point on the ellipse

$PF _{1}+PF _{2}=2a=2\times 5=10$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The curve represented by $Rs \left(\dfrac{1}{z}\right)=C$ is (where $C$ is a constant and $\neq 0$)

  1. Ellipse

  2. Parabola

  3. Circle

  4. Straight line
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { { Re } }\, \, \left( { \frac { 1 }{ z }  } \right) =c \ { { Re } }\, \, \left( { \frac { 1 }{ { x+iy } }  } \right) =c \ { { Re } }\, \, \left( { \frac { { x-iy } }{ { { x^{ 2 } }+{ y^{ 2 } } } }  } \right) =c \ \frac { x }{ { { x^{ 2 } }+{ y^{ 2 } } } } =c \ c\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) -{ x }=0 . \end{array}$


Hence, this is represent circle.

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

The eccentricity of the curve with equation ${ x }^{ 2 }+{ y }^{ 2 }-2x+3y+2=0$ is

  1. $0$
  2. $\sqrt { 2 }$
  3. $1/2$
  4. ${ 1 }/{ \sqrt { 2 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ${x}^{2}+{y}^{2}-2x+3y+2=0$

$\Rightarrow \left({x}^{2}-2x\right)+\left({y}^{2}+3y\right)+2=0$
$\Rightarrow \left({x}^{2}-2x+1-1\right)+\left({y}^{2}+2\times 1\times \dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}\right)+2=0$
$\Rightarrow {\left(x-1\right)}^{2}-1+{\left(y+\dfrac{3}{2}\right)}^{2}-\dfrac{9}{4}+2=0$
$\Rightarrow {\left(x-1\right)}^{2}+{\left(y+\dfrac{3}{2}\right)}^{2}=-2+1+\dfrac{9}{4}=\dfrac{5}{4}$
Divide both sides by ${\left(\sqrt{\dfrac{5}{4}}\right)}^{2}$ we get
$\dfrac{{\left(x-1\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}+\dfrac{{\left(y+\dfrac{3}{2}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}$  is an ellipse where $a=\sqrt{\dfrac{5}{4}}$ and  $b=\sqrt{\dfrac{5}{4}}$
Eccentricity $e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}}=\sqrt{1-1}=0$


Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A curve is such that the midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line $y=x$. If the curve passes through $(1,0)$, then the curve is

  1. $2y=x^2-x$
  2. $y=x^2-x$
  3. $y=x-x^2$
  4. $y=2(x-x^2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $P\left(x,y\right)$ be a point on the curve then equation of the tangent is $Y-y=\dfrac{dy}{dx}\left(X-x\right)$

Given that tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis
$\Rightarrow\,x-$coordinate$=0$

$\Rightarrow\,X=0$

$\Rightarrow\, Y-y=\dfrac{dy}{dx}\left(X-x\right)$ becomes

$\Rightarrow\, Y-y=\dfrac{dy}{dx}\left(0-x\right)$ 

$\Rightarrow\, Y=y-x\dfrac{dy}{dx}$

$\therefore\,A=\left(0,y-x\dfrac{dy}{dx}\right)$

Given that  midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis, lies on the line $y=x$

$\therefore\,$Midpoint of the line $AP$ lies on the line $y=x$

Midpoint of the line $AP=\left(\dfrac{x+0}{2},\,\dfrac{y+y-x\dfrac{dy}{dx}}{2}\right)$ lies on  the line $y=x$

$\therefore\,x-$coordinate$=y-$coordinate

$\Rightarrow\,\dfrac{x+0}{2}=\dfrac{2y-x\dfrac{dy}{dx}}{2}$

$\Rightarrow\,x=2y-x\dfrac{dy}{dx}$ is a linear differential equation.
$\dfrac{dy}{dx}-\dfrac{2}{x}y=-1$

Integrating factor is $={e}^{\int{p\,dx}}={e}^{\int{\dfrac{-2}{x}\,dx}}={e}^{-\ln{x}}=\dfrac{1}{{x}^{2}}$

Now, $\dfrac{1}{{x}^{2}}\times\dfrac{dy}{dx}-\dfrac{1}{{x}^{2}}\times \dfrac{2}{x}y=\dfrac{-1}{{x}^{2}}$

$\Rightarrow\,\dfrac{1}{{x}^{2}}\dfrac{dy}{dx}-\dfrac{2}{{x}^{3}}y=\dfrac{-1}{{x}^{2}}$ 

$\Rightarrow\,\dfrac{1}{{x}^{2}}dy-\dfrac{2}{{x}^{3}}ydx=\dfrac{-dx}{{x}^{2}}$ 

Integrating both sides,we get

$\Rightarrow\,d\left(\dfrac{y}{{x}^{2}}\right)=d\left(\dfrac{1}{x}\right)$

$\Rightarrow\,d\left(\dfrac{y}{{x}^{2}}\right)=d\left(\dfrac{1}{x}\right)$

$\Rightarrow\,\dfrac{y}{{x}^{2}}=\dfrac{1}{x}+c$ is the required curve.

The curve passes through $\left(1,0\right)$

$\Rightarrow\,0=1+c$

$\Rightarrow\,c=-1$

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

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

$\Rightarrow\,y=x-{x}^{2}$ is the required equation of the curve passing through $\left(1,0\right)$
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Consider a curve $a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=1$ and a point $P$ not on the curve. A line drawn from the point $P$ intersect the curve ar point $Q$ and $R$. If the product $PQ.PR$ is independent of the slope of the line, then the curve is

  1. An ellipse

  2. A hyperbola

  3. A circle

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

Let the coordinates of point$P$ be $\left( { x } _{ 1 },{ y } _{ 1 } \right). $

Equation of any line through $P$ can be written as $\displaystyle \frac { x-{ x } _{ 1 } }{ \cos { \theta  }  } =\frac { y-{ y } _{ 1 } }{ \sin { \theta  }  } =r$    ...(1)
$\Rightarrow x={ x } _{ 1 }+r\cos { \theta  } ,y={ y } _{ 1 }+r\sin { \theta  } .$

Coordinates of any point an (1) is of the form $\left( { x } _{ 1 }+r\cos { \theta  } ,{ y } _{ 1 }+r\sin { \theta  }  \right) .$ 
This point will lie on ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }=1$ if
$a\left( { x } _{ 1 }+r\cos { \theta  }  \right) ^{ 2 }+2h\left( { x } _{ 1 }+r\cos { \theta  }  \right) \left( { y } _{ 1 }+r\sin { \theta  }  \right) +b{ \left( { y } _{ 1 }+r\sin { \theta  }  \right)  }^{ 2 }-1=0$
$\Rightarrow { r }^{ 2 }\left( a\cos ^{ 2 }{ \theta  } +2h\cos { \theta  } \sin { \theta  } +b\sin ^{ 2 }{ \theta  }  \right) +2\left[ { x } _{ 1 }\left( a\cos { \theta  } +h\sin { \theta  }  \right) +{ y } _{ 1 }\left( h\cos { \theta  } +b\sin { \theta  }  \right)  \right]$
$ +{ ax } _{ 1 }^{ 2 }+2{ hx } _{ 1 }{ y } _{ 1 }+{ by } _{ 1 }^{ 2 }-1=0$     ...(2)
Let $PQ={ r } _{ 1 }$  and $PR={ r } _{ 2 }.$ 
Then ${ r } _{ 1 },{ r } _{ 2 }$ are the roots of (2).
$\displaystyle \therefore PQ:PR={ r } _{ 1 }{ r } _{ 2 }=\frac { { ax } _{ 1 }^{ 2 }+2{ hx } _{ 1 }{ y } _{ 1 }+{ by } _{ 1 }^{ 2 }-1 }{ a\cos ^{ 2 }{ \theta  } +2h\cos { \theta  } \sin { \theta  } +b\sin ^{ 2 }{ \theta  }  } .$
We know rewrite the denominator.
We have$D=a\cos ^{ 2 }{ \theta  } +2h\cos { \theta  } \sin { \theta  } +b\sin ^{ 2 }{ \theta  } .\\ $
$\displaystyle =\frac { 1 }{ 2 } \left[ \left( a+b \right) +\left( a-b \right) \cos { 2\theta  }  \right] +h\sin { 2\theta  } $
$\displaystyle =\frac { a+b }{ 2 } +\frac { 1 }{ 2 } \left( a-b \right) \cos { 2\theta  } +h\sin { 2\theta  } $
Put $\displaystyle \frac { 1 }{ 2 } \left( a-b \right) =k\sin { \alpha  } ,h=k\cos { \alpha  } .$
$\displaystyle \Rightarrow k=\sqrt { { \left( \frac { a+b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 } } $ and $\displaystyle \tan { \alpha  } =\frac { a-b }{ 2h } $
$\displaystyle \therefore D=\frac { 1 }{ 2 } \left( a+b \right) +\sqrt { { \left( \frac { a-b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 } } \sin { \left( 2\theta +\alpha  \right)  } $
Thus, $\displaystyle PQ.PR=\frac { { ax } _{ 1 }^{ 2 }+2{ hx } _{ 1 }{ y } _{ 1 }+{ by } _{ 1 }^{ 2 }-1 }{ \frac { 1 }{ 2 } \left( a+b \right) +\sqrt { { \left( \frac { a-b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 } } \sin { \left( 2\theta +\alpha  \right)  }  } $
For  this to be independent of $\theta$ we must have $\displaystyle { \left( \frac { a-b }{ 2 }  \right)  }^{ 2 }+{ h }^{ 2 }=0\Rightarrow a=b$ and $n=0.$
But this to be condition for the given curve to represent a circle.  

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The differential equation of the curve such that the ordinates of any point is equal to the corresponding subnormal at that point is

  1. a linear equation

  2. a non-homogeneous equation

  3. an equation with separable variable

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The differential equation of the curve such that the ordinate of the any point is equal to the  corresponding subnormal at that point is a linear equation.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Curves $a{ x }^{ 2 }+2hxy-2gx-2fy+c=0$ and $a'{ x }^{ 2 }-2hxy+(a'+a-b){ y }^{ 2 }-2g'x-2f'y+c=0\quad $ intersect at four concyclic points $A,B,C$ and $D$. If $P$ is the point $\left( \cfrac { g'+g }{ a'+a } ,\cfrac { f'+f }{ a'+a }  \right) $, then which of the following is/are true

  1. $P$ is also concyclic with points $A,B,C,D$
  2. $PA,PB,PC$ in G.P
  3. ${ PA }^{ 2 }+{ PB }^{ 2 }+{ PC }^{ 2 }=3{ PD }^{ 2 }\quad $
  4. $PA,PB,PC$ in AP
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of two conics at four concyclic points is a standard result in geometry. The point P defined by the given ratios is the center of the circle passing through these four points.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The line $\mathrm{l}\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n}=0$ intersects the curve $\mathrm{a}\mathrm{x}^{2}+2\mathrm{h}\mathrm{x}\mathrm{y}+\mathrm{b}\mathrm{y}^{2}=1$ at $\mathrm{P}$ and $\mathrm{Q}$. The circle with $\mathrm{P}\mathrm{Q}$ as diameter passes through the origin then $\displaystyle \frac{l^{2}+m^{2}}{n^{2}}=$

  1. $a + b$
  2. $(\mathrm{a}+\mathrm{b})^{2}$
  3. $\mathrm{a}^{2}+\mathrm{b}^{2}$
  4. $\mathrm{a}^{2}-\mathrm{b}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$lx+my+x=0$
$\displaystyle y=\frac{-x-lx}{m}$ ---1
$\therefore $$ax^{2}+2hxy+by^{2}=1$
Put $\displaystyle y=\frac{-x-lx}{m}$ in the above equation
$\displaystyle ax^{2}+2hx(\frac{-h-lx}{m})+b(\frac{h+lx}{m})^{2}=1$
$\displaystyle ax^{2}-\frac{2xhx}{m}-\frac{2hlx^{2}}{m}+\frac{bx^{2}}{m^{2}}+\frac{bl^{2}x^{2}}{m^{2}}+\frac{2bxlx}{m}=1$
$\displaystyle (a-\frac{2hl}{m}+\frac{bl^{2}}{m^{2}})x^{2}+(\frac{2bxl}{m}-\frac{2xh}{m})x+\frac{bx^{2}}{m^{2}}-1=0$
$\displaystyle \therefore $$x _{2}+x _{2}=\displaystyle \dfrac{\dfrac{2xh-2bxl}{m}}{a-\dfrac{2hl}{m}+\dfrac{bl^{2}}{m^{2}}}$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

Let P be the point (1, 0) and Q a point on the curve ${ y }^{ 2 }=8x$. The locus of mid point of PQ is-

  1. ${ y }^{ 2 }-4x+2=0$
  2. ${ y }^{ 2 }+4x+2=0$
  3. ${ x }^{ 2 }+4y+2=0$
  4. ${ x }^{ 2 }-4y+2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P=(1,0), Q=(h,k)$


$k^2=8h$


Let $(\alpha, \beta)$ be the mid-point of PQ.

$\alpha =\dfrac{h+1}{2}, \beta =\dfrac{k+0}{2}$

$2\alpha-1=h, 2\beta=k$

$(2\beta)^2=8(2\alpha-1)$

$\beta^2=4\alpha-2$

$\implies y^2-4x+2=0$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

A man of height 1.8 metre is moving away from a lamp post at the  rate of 1.2 m/sec . If the height of the lamp post be 4.5 metre , then the rate at which the shadow of the  man is lengthening is 

  1. $0.4 m/sec$
  2. $0.8m/sec$
  3. $1.2m/sec$

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

Let x be the distance of the man from the post and y be the length of the shadow. By similar triangles, y/1.8 = (x+y)/4.5. Simplifying gives 4.5y = 1.8x + 1.8y, so 2.7y = 1.8x, or y = (2/3)x. The rate of change dy/dt = (2/3) * dx/dt = (2/3) * 1.2 = 0.8 m/sec.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The normal to the curve x$^2$ = 4y passing (1,2) is 

  1. x + y = 3

  2. x - y = 3

  3. x + y = 1

  4. x - y = 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$x^2=4y$

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

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

$\dfrac{dy}{dx} _{h,k}=\dfrac{h}{2}$

$\dfrac{-1}{\dfrac{dy}{dx} _{h,k}}=-\dfrac{2}{h}$

Equation of normal

$(y-k)=-\dfrac{2}{h}(x-h)$

Given point, $(1,2)$

$(2-k)=-\dfrac{2}{h}(1-h)$

$k=2+\dfrac{2}{h}(1-h)$

$\therefore k=\dfrac{h^2}{4}$

$\Rightarrow \dfrac{h^2}{4}2+\dfrac{2}{h}(1-h)$

upon solving, we get,

$h=2,k=1$

Hence, the equation of normal is 

$(y-1)=\dfrac{-2}{2}(x-2)$

$y-1=-x+2$

$x+y=3$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The number of normals that can be drawn to the curve $\displaystyle 4x^{2} + 9y^{2} = 36$ from an external point, in general, is

  1. $1$
  2. $3$
  3. $4$

  4. infinite
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given ellipse may be written as, $\displaystyle \cfrac{x^{2}}{9} + \cfrac{y^{2}}{4} = 1$
$\Rightarrow a^2=9, b^2=4$
Thus general equation of normal to ellipse with slope $m$ is given by,
$y = mx-\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}=mx-\cfrac{5m}{\sqrt{9+4m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{25m^2}{9+4m^2}$
$\Rightarrow (y _1-mx _1)^2(9+4m^2)=25m^2$
Clearly this is a polynomial of degree four so maximum number of normal that can be drawn from point $P$ (any external point) to the ellipse is $4$ corresponding to four roots of $m.$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Find where the line $\displaystyle 2x+y=3$ cuts the curve $\displaystyle 4x^{2}+y^{2}=5.$ Obtain the equations of the normals at the points of intersection and determine the co-ordinates of the point where these normals cut each other.

  1. $\displaystyle \left ( -1, \frac{1}{2} \right )$
  2. $\displaystyle \left ( 1, \frac{1}{2} \right )$
  3. $\displaystyle \left ( -1, \frac{-1}{2} \right )$
  4. $\displaystyle \left ( 1, \frac{-1}{2} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle P\left ( \frac{1}{2}+, 2 \right ), Q(1, 1)$ 
Tangents at P and Q are $\displaystyle 4x\cdot \frac{1}{2}+y\cdot 2=5$ and $\displaystyle 4x\cdot 1+y\cdot 1=5$ or $\displaystyle 2x+2y=5$ and $\displaystyle 4x+y=5$ 
Hence normals are $\displaystyle 2x-2y+3=0,$ $\displaystyle x-4y+3=0$
 They intersect at $\displaystyle \left ( -1, \frac{1}{2} \right )$