Mathematics

Calculus and Analytic Geometry

97 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 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 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 $\displaystyle \left ( -1, \frac{1}{2} \right )$

Multiple choice position of point wrt ellipse ellipse maths

The value of $\alpha$ for which the point $(\alpha,\alpha+2)$ is an interior point of smaller segment of the curve $x^{2}+y^{2}-4=0$ made by the chord of the curve whose equation is $3x+4y+12=0$ is

  1. $\left(-\infty,\dfrac {-20}{7}\right)$
  2. $(-2,0)$
  3. $\left(-\infty,\dfrac {20}{7}\right)$
  4. $\alpha\ \epsilon\ \phi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice position of point wrt ellipse ellipse maths

Let a curve satisfying the differential equation $y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$ which passes through $(1, 1)$. If the curve also passes through $(k, 2)$, then value of k is?

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

$y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$
$\Rightarrow \dfrac{dx}{dy}+\dfrac{x}{y^2}=\dfrac{1}{y^3}$
Integrating factor (I.F.)$=e^{-\dfrac{1}{y}}$
Now $x.e^{-\dfrac{1}{y}}=\displaystyle\int e^{-\dfrac{1}{y}}\dfrac{1}{y^3}dy$
Put $-\dfrac{1}{y}=y$
$x.e^t=\displaystyle\int e^t(-t)dt$
$\Rightarrow x.e^t=-(t.e^t-e^t)+c$
$\Rightarrow e^{-\dfrac{1}{y}}=e^{-\dfrac{1}{y}}\left(1+\dfrac{1}{y}\right)+c$
$\Rightarrow x=1+\dfrac{1}{y}+c.e^{\dfrac{1}{y}}$
it passes through point $(1, 1)$
$\therefore c=-\dfrac{1}{e}$
Equation of curve is
$x=1+\dfrac{1}{y}-e^{\dfrac{1}{y}-1}$
It passes through $(k, 2)$
$\therefore k=1+\dfrac{1}{2}-e^{-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}$.

Multiple choice position of point wrt ellipse ellipse maths

The minimum distance of origin from the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is $(a>0,b>0)$

  1. a-b

  2. a+b

  3. 2a+2b

  4. 2(a-b)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $(a sec\theta,b cosec \theta)$ be a point on the curve, then its diastance  from the origin
$=\sqrt{a^2 sec^2 \theta + b^2 cosec ^2 \theta}$
$\therefore f(\theta)=a^2 sec^2 \theta +b^2 cosec ^2 \theta$
$=a^2+a^2 tan ^2 \theta + b^2 + b^2 cot ^ 2 \theta$
$=a^2+b^2 + a^2 tan ^2 \theta + b^2 cot ^2 \theta$
$\geq a^2+b^2+2 \sqrt{a^2 b^2}=(a+b)^2$
$\therefore$ minimum value of $f(\theta)=(a+b)^2$
$\therefore$ minimum value of $\sqrt{a^2 sec^2 \theta + b^2 cosec^2 \theta }$ is $a+b$

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Tangent at a point ${P _1}$ (other than (0, 0) on the curve $y = {x^3}$ meets the curve again at ${P _2}$. The tangent at ${P _2}$ meets the curve again at ${P _3}$ and so on. Show that the abscissae of ${P _1},{P _2},..........,{P _n}$ form a G.P. Also find the ratio $\left[ {area\,\left( {\Delta {P _1}.{P _2}.{P _3}} \right)/area\,\left( {\Delta {P _2}{P _3}{P _4}} \right)} \right].$

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{1}{8}$
  4. $\dfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the curve y = x^3, the tangent at x1 meets the curve at x2 = -2x1. This forms a geometric progression with common ratio -2. The area of the triangle formed by points on the curve scales with the coordinates, leading to a ratio of 1/16 for consecutive triangles.

Multiple choice

What is the equation of the tangent line to the curve $y = x^3 - 2x^2 + x - 1$ at the point $(1, -1)$?

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

The equation of the tangent line to a curve at a point is given by the formula $y - y_1 = m(x - x_1)$, where $m$ is the slope of the tangent line and $(x_1, y_1)$ is the point of tangency. In this case, the slope of the tangent line is $m = f'(1) = 3(1)^2 - 2(1) + 1 = 2$. So the equation of the tangent line is $y - (-1) = 2(x - 1)$, which simplifies to $y = 3x - 4$.

Multiple choice

What is the Picard group of the elliptic curve $y^2 = x^3 + x + 1$?

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

The Picard group of an elliptic curve is a group that is related to the number of linearly independent holomorphic differentials on the curve. In this case, the elliptic curve $y^2 = x^3 + x + 1$ has one linearly independent holomorphic differential, so its Picard group is $\mathbb{Z}$.