MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 3

Equation of the tangent to the parabola

$y^{2}=4x+5$ which is parallel to the line

$y= 2x + 7$ is

0%
$y = 2x + 3$
0%
$y = 2x- 3$
0%
$y = 2x + 5$
0%
$y = 2x - 5$

Explanation

Differentiating the equation given,we obtain

$2{y}'y=2$

${y}'=2$

$\dfrac{2}{y}=2$

$y=1$

$x=-1$

$(y-1)=2(x+1)$

$y=2x+3$

For the parabola

$y^{2}=8x$ , tangent and normal are drawn at

$P(2, 4)$ which meet the axis of the parabola in

$A$ and

$B$ , then the length of the diameter of the circle through

$A, P, B$ is

0%
$2$
0%
$4$
0%
$8$
0%
$6$

Explanation

Considering the equation of the parabola

$y^{2}=8x$

The axis of symmetry is the positive

$x-$ axis.

By, differentiating with respect to

$x$ , we get

$2y.y'=8$

$y'=\dfrac{4}{y}$

Hence,

$y'_{(2,4)}=1$

Thus slope of the tangent at the point of contact

$P(2,4)$ is

$1$ while that of normal is

$-1$ .

Hence, equation of the tangent will be

$\dfrac{y-4}{x-2}=1$

$-x+y=2$ ...(i)

Hence, the tangent meets the x axis at

$(-2,0)$

Therefore,

$A=(-2,0)$ .

Equation of normal will be

$\dfrac{y-4}{x-2}=-1$

$x+y=6$ ...(ii)

Hence the normal meets the x axis at

$(6,0)$ .

Thus,

$B=(6,0)$

Therefore the three points through which the circle passes are

$(-2,0),(6,0),(2,4)$

Now let the equation of the circle be

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Therefore

$(-2-h)^{2}+k^{2}=r^{2}$

$\rightarrow (2+h)^{2}+k^{2}=r^{2}$

$(6-h)^{2}+k^{2}=r^{2}$

Subtracting {ii} from {i}, we get

$2h-4=0$

$h=2$

Therefore the equation of the circle reduces to

$(x-2)^{2}+(y-k)^{2}=r^{2}$
Now

$(2-2)^{2}+(4-k)^{2}=r^{2}$

$\rightarrow (4-k)^{2}=r^{2}$

$(6-2)^{2}+k^{2}=r^{2}$

$\rightarrow 16+k^{2}=r^{2}$

Subtracting i from ii, we get

$16+k^{2}-(k-4)^{2}=0$

$16+(2k-4)(4)=0$

$4+2k-4=0$

$k=0$

Thus the centre of the circle lies at

$C=(2,0)$ .

Hence the radius of the circle is

$CA=CP=CB$
Now

$CA=\sqrt{(2-(-2))^{2}+0^{2}}$

$=2-(-2)$

$=4$

$=r$

Hence, the diameter of the circle is

$8$ units.

If the curves

$y=x^{2}-1,\ y=8x-x^{2}-9$ touch each other at (2, 3) then equation of the common tangent is

0%
$4x-y=5$
0%
$4x+y=5$
0%
$x-4y=5$
0%
$x+4y=14$

Explanation

$y_{1}^{1}=2x$

$y_{2}^{1}=8-2x$
at the tangent point slope will be
some.

$2x=8-2x$

$4x=8$

$x=2$

$y=3$

$y^{1}=4$

$(y-3)=4(x-2)$

$y=4x-5$

The point on the hyperbola

$y = \dfrac {x - 1}{x + 1}$ at which the tangents are parallel to

$y = 2x + 1$ are

0%
$(0, -1)$ only
0%
$(-2, 3)$ only
0%
$(0, -1)$ , $(-2, 3)$
0%
$(-2, 3)$ , $(5, 4)$

Explanation

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

${y}'=2 (\therefore$ parallel to

$y =2x+1)$

$\dfrac{2}{(x+1)^{2}}=2$

$(x+1)^{2}=1$

$x=0$ or

$x=-2$
Using the equation of hyperbola, when

$x=0$ ,

$y=-1$ .
when

$x=-2$ ,

$y=3$

Assertion(A): The tangent to the curve

$y=x^{3}-x^{2}-x+2$ at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.

0%
Both A and R are true R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$y^{1}=3x^{2}-2x-1$

$y^{1}|_{x=1}=0=m$

$m=0$

$\therefore$ the curve is parallel to x-axis.

Assertion A: The curves

$x^{2}=y,\ x^{2}=-y$ touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.

0%
Both A and R are true R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$y'=2x$

$y'|_{x=0}=0=m_{1}$

$y'=-2x$

$y'|_{x=0}=0=m_{2}$

$m_{1}=m_{2}$

$\therefore$ both the curve touches each other

Observe the following statements for the curve

$y=2.e^{\frac{-x}{3}}$
I : The slope of the tangent to the curve where it meets y-axis is

$\displaystyle \frac{-2}{3}$
II:The equation of normal to the curve where it meets y-axis is

$3x+2y+4=0$ .
Which of the above statement is correct

0%
only I
0%
only II
0%
both I and II
0%
neither I nor II

Explanation

$y'=\cfrac{-2}{3}e^{-x}$

$y'|_{x=0}=\frac{-2}{3}$

$y=2,\ x=0$

$(y-2)=\dfrac{3}{2}(x-0)$

$2y-3x-4=0$
only 2 is true

Match the points on the curve

$2y^{2}=x+1$ with the slope of normals at those points and choose

the correct answer.

Point	Slope of normal
I : $(7, 2)$	$a){-4\sqrt{2}}$
II: $(0, \displaystyle \frac{1}{\sqrt{2}})$	$b) -8$
III : $(1, 1)$	$c) -4$
IV: $(3, \sqrt{2})$	$d){-2\sqrt{2}}$

0%
$i-b,ii- d,iii- c,iv- a$
0%
$i-b,ii- a,iii- d,iv- c$
0%
$i-b,ii- c,iii- d,iv- a$
0%
$i-b,ii- d,iii- a,iv- c$

Explanation

$\displaystyle y'=\frac{1}{4y}=m$

Slope of normal

$m'=-4y$

For given points slope is

(1)

$m'_{1}=-4\times 2 ==-8$
(2)

$m'_{2}=-4\times \dfrac{1}{\sqrt 2}=-2\sqrt{2}$
(3)

$m'_{3}=-4\times 1=-4$
(4)

$m'_{4}=-4\times {\sqrt 2}=-4\sqrt{2}$

Option A is correct answer

The equation of the tangent to the curve

$y=e^{-|x|}$ at the point where the curve cuts the line

$x=1$ is

0%
$x+y=e$
0%
$e(x+y)=1$
0%
$y+ ex=1$
0%
$x+ey=2$

Explanation

$y=e^{-x}$

$y'=\dfrac{-1e^{-x}}{e^{-x}}$

$y'=-e^{-1} \ \ at \ \ x=1$

$y=e^{-1} \ \ at \ \ x=1$

$(y-e^{-1})=-e^{-1}(x-1)$

$x+ey=2$

The equation of the normal at

$x$

$=$

$2a$ for the curve

$\displaystyle y=\frac{8a^{3}}{4a^{2}+x^{2}}$ is

0%
$2x-y=3a$
0%
$2x+y=2a$
0%
$x+2y=6a$
0%
$x+y=a$

Explanation

Equation of normal at

$x=2a$ for

$y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ x }^{ 2 } }$

$\Rightarrow x=2a$

$\Rightarrow y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ (2a) }^{ 2 } } =a$

$y'=\dfrac { dy }{ dx } =\dfrac { -8{ a }^{ 3 }(2x) }{ { \left( 4{ a }^{ 2 }+{ x }^{ 2 } \right) }^{ 2 } } ,x=2a\Rightarrow y'=\dfrac { -8{ a }^{ 3 }\times 2\times 2a }{ { \left( 4{ a }^{ 2 }+4{ a }^{ 2 } \right) }^{ 2 } }$

$=\dfrac { -8{ a }^{ 3 }\times 4a }{ 64{ a }^{ 4 } }$

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

Slope of normal

$=\dfrac { -1 }{ y' } =2$

$\Rightarrow y-a=2(x-2a)$

$2x-4a=y-a$

$\therefore 2x-y=3a$ is a equation of normal.

The distance of the origin from the normal to the curve

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

$x=0$ is

0%
$\displaystyle \frac{2}{5}$
0%
$\displaystyle \frac{2}{\sqrt{5}}$
0%
$2\sqrt{5}$
0%
$5\sqrt{2}$

Explanation

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

$y=1$ at

$x=0$

$y'=2$

$m$ of normal

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

$(y-1)=\dfrac{-1}{2}(x-0)$

$x+2y-2=0$

distance from origin

$d=\left |\dfrac{x_{1}+2y_{1}-2}{\sqrt{1^{2}+2^{2}}} \right | (x_{1}y_{1})=(0,0)$

$d=\dfrac{2}{\sqrt{5}}$

Equation of the tangent line to

$y=be^{\frac{-x}{a}}$ where it crosses y-axis is

0%
$ax+ by=1$
0%
$\displaystyle \frac{x}{a}+\frac{y}{b}=1$
0%
$\displaystyle \frac{x}{b}+\frac{y}{a}=1$
0%
$ax- by=1$

Explanation

$y=b e^{\frac{-x}{a}} at x=0$

$y=b$

$y'=\dfrac{-b e^{-x}}{a}$

$y'=\dfrac{-b }{a} \ \ (at \ \ x=0)$

$(y-b )=\dfrac{-b }{a}(x-0)$

$\dfrac{x}{a}+\dfrac{y}{b }=1$

The portion of the tangent to xy

$=a^{2}$ at any point on it between the axes is

0%
Trisected at that point
0%
bisected at that point
0%
constant
0%
with ratio $1 : 4$ at the point

Explanation

Let xy tangent at

$(a,a)$

$xy'+y=0$

$y^{1}=\cfrac{-y}{x}$

$y'=-1$

$(y-a)=-1(x-a)$

Bisected by that point.

$y' \ \ at \ \ B =\dfrac{-y}{x}$

$=\dfrac{-ay^{2}}{ty^{2}}$

$(y-\dfrac{a^{2}}{t})=\dfrac{-a^{2}}{t^{2}}(x-t)$

$x=0 , y=\dfrac{2a^{2}}{t}$

$y=0, x=2t$

$(t,\dfrac{a^{2}}{t})$ to the mid point of (2+10) and

$(0,\dfrac{2a^{2}}{t})$ .
H.P.

Area of the triangle formed by the normal to the curve

$x=e^{\sin y}$ at (1, 0) with the coordinate axes is

0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{3}{4}$
0%
$1$

The arrangment of the slopes of the normals to the curve

$y=e^{\log(cosx)}$ in the ascending order at the points given below.

$A) \displaystyle x=\frac{\pi}{6}, B) \displaystyle x=\frac{7\pi}{4}, C)x=\frac{11\pi}{6}, D)x=\frac{\pi}{3}$

0%
$C, B, D, A$
0%
$B, C, A, D$
0%
$A, D, C, B$
0%
$D, A, C, B$

Explanation

We have,

$y = e^{\log \cos x} =(\cos x)^{\log e}[\because a^{\log_bc} =c^{\log_ba}] =\cos x$

$\Rightarrow \dfrac{dy}{dx} =-\sin x=m$ (say)
Thus slope of normal to the given curve at any point is,

$m'=- \dfrac{1}{m}=\dfrac{1}{\sin x}$
Now,

$m'_{\frac{\pi}{6}}=2, m'_{\frac{7\pi}{4}}=-\sqrt{2}, m'_{\frac{\pi}{6}}=-2,m'_{\frac{\pi}{6}}=\dfrac{\sqrt{3}}{2}$
Clearly ascending order is,

$C,B,D,A$
Note: Given function is not defined where

$\cos x<0$

lf the normal at the point

$p(\theta)$ of the curve

$x^{\tfrac{2}{3}}+y^{\tfrac{2}{3}}=a^{\tfrac{2}{3}}$ passes through the origin then

0%
$\theta =\dfrac {\pi}3$
0%
$\theta =\dfrac {\pi}6$
0%
$\theta =\dfrac {\pi}4$
0%
$\theta =\dfrac {\pi}2$

Explanation

$x=a\ \cos^{3}\theta$

$y=a\ \sin^{3}\theta$

$y^{1}=\dfrac{a\ 3\ \sin^{2}\theta\ \cos\ \theta}{a\ 3\ \cos^{2}\theta-\cos\ \theta}$

$\dfrac{-\sin\ \theta}{\cos\ \theta}(y-a\ \sin^{3}\theta)=\dfrac{\cos\ \theta}{\sin\ \theta}(x-a\ \cos^{3}\ \theta)$

$(0, 0)$ satisfies the eq

$^{n}$

$\sin^{4}\ \theta=\cos^{4}\ \theta$

$\theta=\dfrac {\pi}4$

The equations of the tangents at the origin to the curve

$y^{2}=x^{2}(1+x)$ are

0%
$y=\pm x$
0%
$y=\pm 2x$
0%
$y=\pm 3x$
0%
$x=\pm 2y$

Explanation

$y=+_-\sqrt{x^{2}+x^{3}}$

$y'=+_-\dfrac{2x+3x^{2}}{2\sqrt{x^{2}+x^{3}}}$

$y'=+_-\dfrac{2+3x}{2\sqrt{1+x}}$

$y'|_{x=0}=+_-1$

$y=+_-x$

The equation of the common normal at the point of contact of the curves

$x^{2}=y$ and

$x^{2}+y^{2}-8y=0$

0%
$x=y$
0%
$x=0$
0%
$y=0$
0%
$x+y=0$

Explanation

$y^{2}+y-8y=0$

$y=0.\ y=7$

$x=0,\ x=\sqrt{7}$

$y^{1}=2x=m_{1}$

$yy^{1}-4y^{1}=-x$

$y^{1}=\frac{x}{4-y}=m_{2}$

$m_{1},\ m_{2}\ at\ (0,0)$

$m_{1}=m_{2}=0$
Slope of the normal

$=\alpha$

$\therefore x=0$ is the required line
cl. y-axis

lf the chord joining the points where

$x= p,\ x =q$ on the curve

$y=ax^{2}+bx+c$ is parallel to the tangent drawn to the curve at

$(\alpha, \beta)$ then

$\alpha=$

0%
$2pq$
0%
$\sqrt{pq}$
0%
$\displaystyle \frac{p+q}{2}$
0%
$\displaystyle \frac{p-q}{2}$

Explanation

$y'=2ax+b$

$y'|_{x=b}=(2a\alpha +b)$

$2a\alpha+b=\left ( \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )$

$2a\alpha+b=\dfrac{ap^{2}+bp-aq^{2}-bq}{(p-q)}$

$=\dfrac{a(p-q)(p+q)+b(p-a)}{(p-q)}$

$2a\alpha+b=a(p+q)+b$

$\alpha=\left ( \dfrac{p+q}{2} \right )$

The arrangement of the following curves in the ascending order of slopes of their tangents at the given points.

$A) \displaystyle y=\frac{1}{1+x^{2}}$ at

$x=0$

$B) y=2e^{\frac{-x}{4}},$ where it cuts the y-axis

$C) y= cos(x)$ at

$\displaystyle x=\frac{-\pi}{4}$

$D) y=4x^{2}$ at

$x=-1$

0%
DCBA
0%
ACBD
0%
ABCD
0%
DBAC

Explanation

(A)

$y^{1}=-\frac{2x}{(1+x^{2})}$

$y^{1}|_{x=0}=0$

(B)

$y^{1}=-\frac{2x}{(4}$

$y^{1}|_{x=0}=-\frac{1}{2}$

(C)

$y^{1}=-sin\ x$

$y^{1}|_{x=^{-pi}/_4}=\frac{1}{\sqrt{2}}$

(D)

$y^{1}=8x$

$y^{1}|_{x=-1}=-8$

$(C)>(A)>(B)>(D)$

Observe the following lists for the curve

$y=6+x-x^{2}$ with the slopes of tangents at the given points; I, II, III, IV

Point	Tangent slope
I: $(1, 6)$	a) $3$
II: $(2, 4)$	b) $5$
III: $(-1, 4)$	c) $-1$
IV: $(-2, 0)$	d) $-3$

0%
$a ,b, c ,d$
0%
$b, c, d ,a$
0%
$c, d ,b ,a$
0%
$c, d ,a ,b$

Explanation

$y^{'}=\ \ \ 1-2x$
(A)

$m_{1}=-1$
(B)

$m_{2}=-3$
(C)

$m_{3}=3$
(D)

$m_{4}=5$

lf the tangent to the curve

$2y^{3}=ax^{2}+x^{3}$ at the point

$(a,\ a)$ cuts off intercepts

$\alpha$ and

$\beta$ on the coordinate axes such that

$\alpha^{2}+\beta^{2}=61$ , then

$a$ is equal to

0%
$\pm 30$
0%
$\pm 5$
0%
$\pm 6$
0%
$\pm 61$

Explanation

$2\times 3\times y^{2}y'=2ax+3x^{2}$

$y' = \dfrac{2ax+3x^2}{6y^2}$

$m = \left.y'\right|_{(a,a)}=\dfrac{5}{6}$

Equation of the tangent at

$(a,a)$ with slope,

$m = \dfrac56$

$(y-a)=\dfrac{5}{6}(x-a)$

Intersection points with coordinate axes are,

$\left(0,\dfrac{a}{6}\right)$ and

$\left(-\dfrac{a}{5},0\right)$

So,

$\alpha = \dfrac a6$ and

$\beta = -\dfrac a5$

Given that,

${\alpha}^2 + {\beta}^2 = 61$

$\Rightarrow \dfrac{a^2}{36} + \dfrac{a^2}{25} = 61$

$\Rightarrow a^2 = {30}^2$

$\Rightarrow a = \pm 30$

Assertion(A): If the tangent at any point

$P$ on the curve

$xy = a^{2}$ meets the axes at

$A$ and

$B$ then

$AP : PB = 1 : 1$
Reason(R): The tangent at

$P(x, y)$ on the curve

$X^{m}.Y^{n}=a^{m+n}$ meets the axes at

$A$ and

$B$ . Then the ratio of

$P$ divides

$\overline{AB}$ is

$n : m$ .

0%
Both A and R are true R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

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

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

$y'=-\dfrac{a^{2}}{t^{2}}$

$\left ( y-\dfrac{a^{2}}{t} \right )=\dfrac{-a^{2}}{t^{2}}(n-t)$

$B=(2+10)$

$A=\left ( 0,\ \dfrac{2a^{2}}{t} \right )$

$\dfrac{AP}{BP}=1$

The point on the curve

$\sqrt{x}+\sqrt{y}=2a^{2}$ , where the tangent is equally inclined to the axes, is

0%
$\left ( a^{4}, a^{4} \right )$
0%
$\left ( 0, 4a^{4} \right )$
0%
$\left ( 4a^{4}, 0 \right )$
0%
none of these

Explanation

$y=x$ is the equation of line as it is equally inclined to axes
So let point

$(k,k)$ is the point of tangency
Therefore,

$\sqrt { k } +\sqrt { k } =2{ a }^{ 2 }\\ k={ a }^{ 4 }$
Hence point is

$\left( { a }^{ 4 },{ a }^{ 4 } \right)$

Match List-I with List-II and select the correct answer using the code given below. A B C D

List-I	List-II
a) Equation of tangent to the curve $y=be^{-x/a}$ at $x=0$	1) $x-2y=2$
b) Equation of tangent to the curve $y=x^{2}+1$ at $(1, 2)$	2) $y = 2x$
c) Equation of normal to the curve $y=2x-x^{2}$ at $(2, 0)$	3) $x-y =\pi$
d) Equation of normal to the curve $y= \sin x$ at $x=\pi$	4) $\displaystyle {\frac{x}{a}+\frac{y}{b}=1}$

0%
4 1 2 3
0%
4 2 1 3
0%
1 2 3 4
0%
1 4 3 2

Explanation

(A)

$\displaystyle y'=\frac{-b}{a}e^{-x/a}$

$\displaystyle m=-\frac{b}{a}$

$\displaystyle (y-b)=\frac{-b}{a}(x-0)$

$\displaystyle \frac{x}{a}+\frac{y}{b}=1$

(B)

$y'=2x$

$y'|_{x=1}=2$

$(y-2)=2(x-1)$

$y=2x$

(C)

$y'=2-2x$

$y'|_{x=2}=-2$
Slope of normal

$\displaystyle =\frac{1}{2}$

$\displaystyle (y-0)=\frac{1}{2}(x-2)$

$2y=x-2$

(D)

$y'=\cos x$

$y'|_{x={\pi}}=-1$
Slope of normal =1

$(y-0)=1(x-\pi)$

$x-y=\pi$

Area of the triangle formed by the tangent, normal at

$(1, 1)$ on the curve

$\sqrt{x}+\sqrt{y}=2$ and the y axis is (in sq. units)

0%
$1$
0%
$2$
0%
$\displaystyle \frac{1}{2}$
0%
$4$

Explanation

$\dfrac{1}{\sqrt{x}}+\dfrac{y^{1}}{\sqrt{y}}=0$

$y'=-\sqrt{\dfrac{y}{x}}$

$y'=-1\ \ at\ (n=1,\ y=1)$

$(y-1)=-1(x-1)$

$y=2$

$(y-1)=1(x-1)$

$y=0$

$Area=\dfrac{1}{2}\times 2 \times 1$

$=1$

A curve with equation of the form

$y=ax^{4}+bx^{3}+c+cx+d$ has zero gradient at the point (0,1) and also touches the x-axis at the point (-1, 0) then the values of the x for which the curve has a negative gradient are :

0%
$x > -1$
0%
$x < 1$
0%
$x < -1$
0%
$-1\leq x\leq 1$

Explanation

$\dfrac{dy}{dx}|_{x=0}=4ax^{3}+3bx^{2}+c=0$

$\therefore c=0$

$\dfrac{dy}{dx}<0$

$4ax^{3}+3bx^{2}<0$

$x^{2}(4ax+3b)<0$

$x<-\dfrac{3b}{4a}$

Assertion (A): The points on the curve

$y=x^{3}-3x$ at which the tangent is parallel to

$x$ -axis are

$(1, -2)$ and

$(-1, 2).$
Reason (R): The tangent at

$(x_{1}, y_{1})$ on the curve

$y=f(x)$ is vertical then

$\displaystyle \frac{dy}{dx}$ at

$(x_{1}, y_{1})$ is not defined.

0%
Both A and R are true and R is the correct explanation for A
0%
Both A and R are true but R is not the correct explanation for A
0%
A is true but R is false
0%
A is false but R is true

Explanation

$y'=3x^{2}-3$

$y'=0$

$\therefore$ tangent is parallel to x-axis

$x^{2}=1$

$x =^+_-1$

$y=-2$

$y=2$
y tangent is vertical then

$\dfrac{dy}{dx}$ is not defined.

lf the tangent at any point on the curve

$x^{4}+y^{4}=c^{4}$ cuts off intercepts

$a$ and

$b$ on the coordinate axes, the value of

$a^{-4/3}+b^{-4/3}$ is

0%
$c^{-4/3}$
0%
$c^{-1/2}$
0%
$c^{1/2}$
0%
$c^{{4}/{3}}$

Explanation

Given equation of curve is

$x^4+y^4=c^4$

$\dfrac{dy}{dx}=-\dfrac{x^3}{y^3}$
Slope of tangent at point

$(x_1,y_1)$ on the curve is

$-\dfrac{x_1^3}{y_1^3}$

Let the equation of tangent to curve be

$\dfrac{x}{a}+\dfrac{y}{b}=1$
Since, this line intersects X-axis at

$(a,0)$ and

$(0,b)$

$\dfrac{x}{c^4/x_1^3}+\dfrac{y}{c^4/y_1^3}=1$

$\Rightarrow a=\dfrac{c^4}{x_1^3},b=\dfrac{c^4}{y_1^3}$

$\Rightarrow a^{-4/3}=\dfrac{c^{-16/3}}{x_1^{-4}}, b^{-4/3}=\dfrac{c^{-16/3}}{y_1^{-4}}$

$\Rightarrow a^{-4/3}+b^{-4/3}=c^{-16/3}(x_1^4+y_1^4)$

$\Rightarrow a^{-4/3}+b^{-4/3}=c^{-4/3}$

I. lf the curve

$y=x^{2}+ bx +c$ touches the straight line

$y=x$ at the point

$(1, 1)$ then

$b$ and

$c$ are given by

$1, 1.$
II. lf the line

$Px+ my +n=0$ is a normal to the curve

$xy=1$ , then

$P> 0,\ m <0$ .
Which of the above statements is correct

0%
only I
0%
only II
0%
both I and II
0%
Neither I nor II

Explanation

(I)

$1+b+c=1$

$b+c=0$

$y'=2x+b$

$y'=1$

$2+b=1$

$(b=-1)$

$c=1$
(II)

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

$m=-\dfrac{1}{x^{2}}$

$-\dfrac{P}{m}=\dfrac{x^{2}}{x}$

$-P=x^{2}m$
p and m will be of opposite num.

At origin the curve

$y^{2}=x^{3}+x$

0%
Touches the x-axis
0%
Touches the y-axis
0%
Bisects the angle between the axes
0%
touches both the axes

Explanation

$\dfrac{dy}{dx}=\dfrac{3x^{2}+1}{2y}$

At origin

$x=0,y=0$

$\tan\ \theta=\infty$

$\theta =\dfrac{\pi}{2}$

$\therefore$ Curve touches the y-axis

Observe the following statements
I: If

$p$ and

$q$ are the lengths of perpendiculars from the origin on the tangent and normal at any point on the curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ then

$4p^{2}+q^{2}=1$ .
II: If the tangent at any point

$P$ on the curve

$x^{3}.y^{2}=a^{5}$ cuts the coordinate axes at

$A$ and

$B$ then

$AP : PB = 3 : 2$

0%
only I
0%
only II
0%
both I and II
0%
neither I nor II

Explanation

$x=a\ \cos^{3}\theta$

$y=a\ \sin^{3}\theta$

$y'=-\tan\ \theta$

$(y-sin^{3}\theta)=-\tan\ \theta(x-\cos^{3}\theta)$

$y+n\ \tan\ \theta-\tan\ \theta\ \cos^{3}\theta-\sin^{3}\theta=0$
Perpendicular from origin

$\left | \dfrac{\tan\ \theta\ \cos^{3}\theta+\sin^{3}\theta}{\sqrt{1+\tan^{2}\theta}} \right |=p$

$p_{2}\ \ \sin\ \theta\ \cos\ \theta$

$p=\dfrac{\sin^{2}\theta}{2}$

$(y-\sin 3\theta)=\frac{1}{\tan\ \theta}(x-\cos3\theta)$

$y\ \tan\ \theta-\sin3\theta\ \tan\ \theta-x+\cos3\ \theta=0$
Perpendicular frm the origin

$\left | \dfrac{\cos3\theta-\sin3\theta\ \tan\ \theta}{\sqrt{1+\tan^{2}\theta}} \right |=q$

$q=\cos2\theta$

$4p^{2}+q^{2}=1$

Observe the following statements for the curve

$x = at^{3}$ ,

$y = at^{4}$ at

$t = 1$ .
I : The equation of the tangent to the curve is

$4x-3y- a = 0$
II : The equation of the normal to the curve is

$3x +4y- 7a = 0$
III: Angle between tangent and normal at any point on the curve is

$\displaystyle \frac{\pi}{2}$
Which of the above statements are correct.

0%
I and II
0%
II and III
0%
I and III
0%
I, II, III

Explanation

$\dfrac{dy}{dx}=\dfrac{4at^{3}}{3at^{2}}$

$\dfrac{dy}{dx}=\dfrac{4t}{3}$

$\dfrac{dy}{dx}|_{t=1}=\dfrac{4}{3}$

$(y-a)=\dfrac{4}{3}(x-a)$ (tangent)

$4x-3y-a=0$

$(y-a)=-\dfrac{3}{4}(x-a)$ (Normal)

$4y+3x-7a=0$

Tangent and normal are always at an angle of

${\pi}/{2}$ to each other

Assertion (A): The normal to the curve

$ay^{2}=x^{3}(a\neq 0, x\neq 0)$ at a point

$(x, y)$ on it makes equal intercepts on the axes, then

$\displaystyle x=\frac{4a}{9}$ .
Reason (R): The normal at

$(x_{1}, y_{1})$ on the curve

$y=f(x)$ makes equal intercepts on the coordinate axes, then

$\left.\displaystyle \frac{dy} {dx}\right|_{(x_{1},y_{1})}=1$

0%
Both (A) and (R) are true and (R) is the correct explanation for (A).
0%
Both (A) and (R) are true but (R) is not the correct explanation for (A).
0%
(A) is true but (R) is false.
0%
(A) is false but (R) is true.

Explanation

$ay^{2}=x^{3}$

$2ay\dfrac{dy}{dx}=3x^{2}\Rightarrow\dfrac{dy}{dx}=\dfrac{3x^{2}}{2ay}$
Now, slope of normal

$= -\dfrac{dx}{dy}=-\dfrac{2ay}{3x^{2}}=-\dfrac{2a\times x^{\frac{3}{2}}}{3x^{2}\sqrt{a}}=-\dfrac{2\sqrt{a}}{3\sqrt{x}}$
For the normal to make equal intercepts on the axes, the slope has to be

$\pm1$ .
Hence,

$\dfrac{-2\sqrt{a}}{3\sqrt{x}}=\pm1$
On squaring both sides,

$\dfrac{4a}{9x}=1\Rightarrow x=\dfrac{4a}{9}$ .

Given the curves

$y=f(x)$ passing through the point

$(0, 1)$ and

$y=\displaystyle \int_{-\infty}^{x}{f(t)}$ passing through the point

$\left( 0, \dfrac{1}{2}\right).$ The tangents drawn to both the curves at the points with equal abscissae intersect on the

$x$ - axis. Then the curve

$y=f(x)$ is

0%
$f(x)=x^2+x+1$
0%
$f(x)=\dfrac{x^2}{e^x}$
0%
$f(x)=e^{2x}$
0%
$f(x)=x-e^x$

The point of intersection of the tangents drawn to the curve

$x^{2}y=1-y$ at the points where it is met by the curve

$xy=1-y$ is given by

0%
$(0, -1)$
0%
$(1, 1)$
0%
$(0, 1)$
0%
$(1, \infty)$

Explanation

$x^{2}y=xy$

$y=1,\ x=0$

$y=^{1}/_{2},\ x=1$

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

$\dfrac{dy}{dx}=-2xy^{2}$

$(x=0)$

$\dfrac{dy}{dx}=0$

$(y-1)=x-0$

$y=1$ ------- (1)

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

$x=0$ (By (1) & (2))

$(0, 1)$

The number of tangents to the curve

$x^{3/2}+y^{3/2}=a^{3/2}$ , where the tangents are equally inclined to the axes, is

0%
$2$
0%
$1$
0%
$0$
0%
$4$

Explanation

$\dfrac{dy}{dx}=-1$ (where x is equally inclined)

$\dfrac{3}{2}x^{^{1}/_{2}}+^{3}/_{2}y^{^{1}/_{2}}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}=1$

$-\sqrt{\dfrac{x}{y}}=-1$

$\sqrt{x}=\sqrt{y}$

$x=y$

$x^{^{3}/_{2}}+x^{^{3}/_{2}}=a^{^{3}/_{2}}$

$2x^{^{3}/_{2}}=a^{^{3}/_{2}}$

$x=\dfrac{a}{2^{^{2}/_{3}}}$

$y=\dfrac{a}{2^{^{2}/_{3}}}$

$\therefore$ only one tangent.

The points on the hyperbola

$x^{2}-y^{2}=2$ closest to the point (0, 1) are

0%
$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$
0%
$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$
0%
$(\displaystyle \frac{1}{2},\frac{1}{2})$
0%
$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$

Explanation

Let point be

$(\sqrt 2 sec\theta,\sqrt 2 tan\theta)$ closest to (0, 1)
So distance

$=\sqrt {2 sec^2\theta+(\sqrt 2 tan\theta-1)^2}$

$=\sqrt {2 sec^2\theta+2tan^2\theta-2\sqrt 2 tan\theta+1}$

$=\sqrt {4 tan^2\theta-2\sqrt 2 tan\theta+3}$

$=2\sqrt {tan^2\theta-\frac {tan\theta}{\sqrt 2}+\frac {3}{4}}$

$=2\sqrt {(tan\theta-\frac {1}{2\sqrt 2})^2+\frac {3}{4}-\frac {1}{8}}$
So distance is min when

$tan\theta=\frac {1}{2\sqrt 2}$

$sec\theta=\pm \frac {3}{2\sqrt 2}$

lf the tangent to the curve

$f(x)=x^{2}$ at any point

$(c, f(c))$ is parallel to the line joining points

$(a, f(a))$ and

$(b,f(b))$ on the curvel then

$a,\ c,\ b$ are in

0%
AP
0%
GP
0%
HP
0%
AGP

Explanation

$\partial^{'}M=2s$ at

$(c,\ \partial(c))=2c$

$\dfrac{\partial(a)-\partial(L)}{a-b}=2c$

$\dfrac{a^{2}=b^{2}}{a-b}=2c$

$a+b=2c$

$\therefore\ a,c,b$ are in AP

$A$ curve is given by the equations

$x=\sec^{2}\theta, y= \cot\theta$ . If the tangent at

$P$ where

$\displaystyle \theta=\frac{\pi}{4}$ meets the curve again at

$Q$ , then length

$PQ$ , is

0%
$\displaystyle \frac{\sqrt{5}}{2}$
0%
$\displaystyle \frac{3\sqrt{5}}{2}$
0%
$10$
0%
$20$

Explanation

$y = \cot\theta$

$\Rightarrow \dfrac{dy}{d\theta}=-\text{cosec}^{2}\theta$

$x = \sec^2\theta$

$\Rightarrow \dfrac{dx}{d\theta}=2\sec\theta\cdot \sec\theta\ \tan\theta$

$\left.\dfrac{dy}{dx}\right|_{\theta = \frac{\pi}4}=-\dfrac{\text{cosec}^{2}\theta}{2\ \sec^{2}\theta\ \tan\theta}$

$=\dfrac{-2}{2\times 2\times 1}=-\dfrac{1}{2}$

$y=\dfrac{1}{\tan\theta}$

$x=1+\tan^{2}\theta$

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

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

$\Rightarrow x=2,\ y=1$
Now, at

$(x,y) = (2,1)$ , equation of a line with slope

$-\dfrac12$ , is

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

$\Rightarrow 2y+x=4$

$\Rightarrow y^{2}(4-2y)=y^{2}+1$

$\Rightarrow 4y^{2}-2y^{3}=y^{2}+1$

$\Rightarrow 2y^{3}-3y^{2}+1=0$

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

$\Rightarrow x=2,\ 5$

$PQ = \sqrt{(2-5)^{2}+\left (1+\dfrac{1}{2} \right )^{2}}$

$=\left ( 3^{2}+\dfrac{3^{2}}{4} \right )^{0.5}$

$=\dfrac{3\sqrt{5}}{2}$

Hence, option B.

lf the parametric equation of a curve given by

$x=e^{t}\cos t,\ y=e^{t}\sin t$ , then the tangent to the curve at the point

$t=\dfrac{\pi}{4}$ makes with axis of

$x$ the angle.

0%
$0$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{2}$

Explanation

$at\ t={\pi}/{4}$

$y=e^{t}sint$

$\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)$

$\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)$

$\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}$

$= \alpha$

$\therefore tan^{-1}(\infty)={\pi}/{2}$

The normal to the curve

$\mathrm{x}=\mathrm{a}(1 +\cos\theta ),\ \mathrm{y}=a\sin\theta$ at any point

$\theta$ always passes through the fixed point:

0%
$(a, 0)$
0%
$(0, a)$
0%
$(0, 0)$
0%
$(a, a)$

Explanation

$x=a\left( 1+\cos { \theta } \right) ,y=a\sin { \theta }$
Differentiating w.r.t

$\theta$ , we get

$\displaystyle \frac { dx }{ d\theta } =a\left( -\sin { \theta } \right) ,\frac { dy }{ d\theta } =a\cos { \theta }$

$\displaystyle \therefore \frac { dy }{ dx } =-\frac { \cos { \theta } }{ \sin { \theta } }$
Therefore equation of normal at

$\left( a\left( 1+\cos { \theta } \right) ,\sin { \theta } \right)$ is

$\displaystyle \left( y-a\sin { \theta } \right) =\frac { \sin { \theta } }{ \cos { \theta } } \left( x-a\left( 1+\cos { \theta } \right) \right)$
It is clear that in the given options normal passes through the point

$\left( a,0 \right)$

lf the curve

$y=px^{2}+qx+r$ passes through the point (1, 2) and the line

$y=x$ touches it at the origin, then the values of

$p,\ q$ and

$r$ are

0%
$p=1,q=-1,r=0$
0%
$p=1,q=1,r=0$
0%
$p=-1,q=1,r=0$
0%
$p=1,q=2,r=3$

Explanation

$2=p+q+r$

$r=0$

$\dfrac{dx}{dy}|_{x=0}=1$

$\dfrac{dy}{dx}=2px+q$

$q=1$

$2=p+1+0$

$p=1$

For the curve

$y=3\sin \theta\cos\theta, x=e^{\theta}\sin \theta, 0\leq \theta\leq\pi$ ; the tangent is parallel to

$x$ -axis when

$\theta$ is

0%
$0$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{\pi}{6}$

Explanation

$\dfrac{dy}{dx}=0$

$\dfrac{dy}{d\theta}=d\dfrac{3}{2}(sin2\theta)$

$\dfrac{dy}{d\theta}=d\dfrac{3}{4}(cos2\theta)$ ---- (1)

$\dfrac{dy}{d\theta}=e^{\theta}sin\theta+e^{\theta}cos\theta$ ---- (2)

$\dfrac{dx}{dy}=\dfrac{3}{4}\dfrac{cos2\theta}{e^{\theta}(sin\ \theta+ cos\ \theta)}=0$

$cos2\theta=0$

$2\theta={\pi}/{2}$

$\theta={\pi}/{4}$

If the circle

$x^{2}+y^{2}+2gx+2fy+c=0$ is touched by

$y=x$ at

$P$ such that

$OP=6\sqrt{2},$ then the value of

$c$ is

0%
36
0%
144
0%
72
0%
None of these

Explanation

Let the point of contact be

$x_{1},y_{1}$
However it lies on the line

$y=x$
Hence

$x_{1}=y_{1}$
Applying distance formula, we get

$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$

$2x_{1}^2=72$

$x_{1}=6$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.

$2x+2yy'+2g+2fy'=0$
Now

$y'=1$ since

$y'$ is the slope of the line

$y=x$ .
By substituting, we get

$x+y=-(g+f)$
Now

$x_{1}=y_{1}=6$
Hence

$12=-(g+f)$ ...(i)
Substituting

$x=y=6$ in the equation of the circle, we get

$36+36+2(6)(g+f)+c=0$

$72+12(-12)+c=0$

$c=144-72$

$c=72$

$x+ 4y=14$ is a normal to the curve

$y^2=\alpha x ^3-\beta$ at

$(2,3)$ , then the value of

$\alpha+\beta$ is

0%
$9$
0%
$-5$
0%
$7$
0%
$-7$

Explanation

Given equation of curve is

$y^2=\alpha x ^3-\beta$
Since, it passes through (2,3)

$4=8\alpha -\beta$

$\displaystyle \frac{dy}{dx}=\frac{3\alpha x^2}{2y}$
Slope of tangent at (2,3) is

$2\alpha$
Slope of normal at

$\displaystyle (2,3)= -\frac{1}{2\alpha}$
Given equation of normal is

$x+4y=14$
Slope of normal is

$-\cfrac{1}{4}$

$\Rightarrow \alpha =2$

$\Rightarrow \beta=7$
Hence

$\alpha+\beta =9$

The number of tangents to the curve

$x^{3/2} + y^{3/2}= 2a^{3/2}$ ,

$a>0$ , which are equally inclined to the axes, is

0%
$2$
0%
$1$
0%
$0$
0%
$4$

Explanation

For a tangent to be equally inclined to the axes, the slope of the tangent should be 1.

Thus

$y'=1$ .

Now

$x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}$ .

$\dfrac{3}{2}(\sqrt{x})+\dfrac{3}{2}(\sqrt{y})=0$

$\sqrt{x}=-\sqrt{y}$

$x=y$

$x_{1}=y_{1}$

Substituting int he above equation, we get

$2x^{\frac{3}{2}}=2a^{\frac{3}{2}}$

$x=y=a$ .

Hence the tangent touches the curve at

$(a,a)$ .

Thus we get only one tangent, satisfying that criteria.

$m$ is the slope of a tangent to the curve

$e^y= 1+x^2$ , then

0%
$\left | m \right | > 1$
0%
$m>1$
0%
$m>-1$
0%
$\left | m \right | \leq 1$

Explanation

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

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

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

Now, since

$x^{2}\ge 0$

$\Rightarrow x^{2}+1 \ge 1$

$\Rightarrow \dfrac{1}{ x^{2}+1} \le 1$

$\Rightarrow \dfrac{2x}{ x^{2}+1} \le 1$

$|m| \le 1$

If the circle

$x^2 + y^2 + 2gx + 2fy + c =0$ is touched by y = x at P in the first quadrant, such that

$OP = 6 \sqrt2$ , then the value of

$c$ is

0%
36
0%
144
0%
72
0%
None of these

Explanation

Let the point of contact be

$x_{1},y_{1}$
However it lies on the line

$y=x$
Hence

$x_{1}=y_{1}$
Applying distance formula, we get

$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$

$2x_{1}^2=72$

$x_{1}=6$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.

$2x+2yy'+2g+2fy'=0$
Now

$y'=1$ since

$y'$ is the slope of the line

$y=x$ .
By substituting, we get

$x+y=-(g+f)$
Now

$x_{1}=y_{1}=6$
Hence

$12=-(g+f)$ ...(i)
Substituting

$x=y=6$ in the equation of the circle, we get

$36+36+2(6)(g+f)+c=0$

$72+12(-12)+c=0$

$c=144-72$

$c=72$

The equations of the tangents to the curve

$y = x^4$ from the point (2, 0) not on the curve, are given by

0%
$y=0$
0%
$y-1=5(x-1)$
0%
$\displaystyle y - \frac{{4098}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)$
0%
$\displaystyle y - \frac{{32}}{{243}} = \frac{{80}}{{81}}\left( {x - \frac{2}{3}} \right)$

Explanation

$y=4\left ( \frac{8}{3} \right )^3 (x-2)$

$y-0 = m(x-2)$

$y=m(x-2)$

$m(x_1 -2) = x_1^4$

$4x_1^4 - 8 x_1^3 = x_1^4$ (

$m=4x_1^3$ )

$3x_1^4 = 8x_1^3$

$x_1 =\frac{8}{3}$

$\therefore \left ( y -\frac{4098}{81} \right ) = \frac{2048}{27} \left ( x- \frac{8}{3} \right )$

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 3 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 3 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.