Tangents and its Equations Test 41

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Tangents and its Equations Test 41

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Weekly Quiz Competition

Question 1
1 / -0

The angle made by the tangent at any point on the curve $$x=a(t+\sin { t } \cos { t } ),y=a{ (1+\sin { t } ) }^{ 2 }$$ with x-axis is

A
$$\dfrac { \pi }{ 2 } $$

B
$$\dfrac { \pi }{ 4 } $$

C
$$\pi +\dfrac { t }{ 2 } $$

D
$$\dfrac { \pi }{ 4 } +\dfrac { t }{ 2 } $$

Solution

Given,

$$x=a(t+\sin t \cos t)$$

$$\dfrac{dx}{dt}=\dfrac{d}{dt}a(t+\sin t \cos t)$$

$$\therefore \dfrac{dx}{dt}=a[1+\cos ^2t-\sin ^2t]=a\left(1+\cos \left(2t\right)\right)=2a\cos ^2t$$

$$y=a(1+\sin t)^2$$

$$\dfrac{dy}{dt}=\dfrac{d}{dt}a(1+\sin t)^2$$

$$\therefore \dfrac{dy}{dt}=2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)$$

$$\Rightarrow \dfrac{dy}{dx}$$

$$=\dfrac{2a\left(1+\sin \left(t\right)\right)\cos \left(t\right)}{2a\cos ^2t}$$

$$=\dfrac {1+\sin t}{\cos t}$$

$$=\dfrac{\cos \frac{t}{2}+\sin \frac{t}{2}}{\cos \frac{t}{2}-\sin \frac{t}{2}}$$

$$=\dfrac{1+\tan \frac{t}{2}}{1 -\tan \frac{t}{2}}$$

$$=\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$$

angle made by tangent,

$$\tan \theta =\tan \left ( \dfrac{\pi }{4}+\dfrac{t}{2} \right )$$

$$\Rightarrow \theta =\dfrac{\pi }{4}+\dfrac{t}{2}$$
Question 2
1 / -0

If tangent at any point on the curve $${ y }^{ 2 }=1+{ x }^{ 2 }\ makes\ an\ angle\ \theta $$ with positive direction of the x-axis then

A
$$\left| \tan { \theta } \right| >1$$

B
$$\left| \tan { \theta } \right| <1$$

C
$$\left| \tan { \theta } \right| \ge1$$

D
$$\left| \tan { \theta } \right| \le 1$$

Solution
Question 3
1 / -0

The equation of the normal to the curve $$y=(1+x)^{ y }+\sin { ^{ -1 }(\sin ^{ 2 }{ x)\ at\ x=0\ is } } $$.

A
$$x+y=1$$

B
$$x-y+1=0$$

C
$$x+y=2$$

D
$$2x-y+1=0$$

Solution
Question 4
1 / -0

If the line $$ax+y=c$$, touches both the curves $${x}^{2}+{y}^{2}=1$$ and $${y}^{2}=4\sqrt{2}x$$, then $$\left| c \right| $$ is equal to:

A
$$1/2$$

B
$$2$$

C
$$\sqrt{2}$$

D
$$\cfrac { 1 }{ \sqrt { 2 } } $$

Solution

Tangent to $${y}^{2}=4\sqrt{2}x$$ is $$y=mx+\cfrac { \sqrt { 2 } }{ m } $$ is also tangent to $${x}^{2}+{y}^{2}=1$$

$$\Rightarrow \left| \cfrac { \sqrt { 2 } /m }{ \sqrt { 1+{ m }^{ 2 } } } \right| =1\Rightarrow m=\pm 1$$

Tangent will be $$y=x+\sqrt{2}$$ or $$y=-x-\sqrt{2}$$

compare with $$y=-ax+c$$

$$\Rightarrow a=\pm 1;c=\pm \sqrt { 2 } $$

$$\therefore |c|=\sqrt 2$$
Question 5
1 / -0

The equation of the normal to the curve$$y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0$$ is

A
x+y=1

B
x-y+1=0

C
2x+y=2

D
2x-y+1

Solution

Given,

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

when $$x=0$$

$$y=1^y+0=1$$

$$\therefore (x_1,y_1)=(0,1)$$

$$\dfrac{dy}{dx}=(1+x)^y\left [ \ln (1+x)\dfrac{dy}{dx}+\dfrac{y}{1+x} \right ]+\dfrac{2\sin x\cos x}{\sqrt{1-\sin ^4x}}$$

$$\dfrac{dy}{dx}_{(0,1)}=1[1]+0$$

$$\therefore \dfrac{dy}{dx}_{(0,1)}=1$$

Therefore, slope of normal is $$-1$$

Equation of normal,

$$y-y_1=\dfrac{-1}{\frac{dy}{dx}}(x-x_1)$$

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

$$y-1=-x$$

$$x+y=1$$ is the equation of normal
Question 6
1 / -0

The sum of the length of sub tangent of sub tangent and tangent to the curve
$$x=c\left[ 2cos\theta -log\left( cos\quad ec\theta +cot\theta \right) \right] ,y=csin2\theta \quad at\quad \theta =\frac { \pi }{ 3 } is$$

A
$$\dfrac { c }{ 2 } $$

B
$$2c$$

C
$$\dfrac { 3c }{ 2 } $$

D
$$\dfrac { 5c }{ 2 } $$

Solution

Given,

$$x=c[2\cos \theta −\log (\csc \theta +\cot \theta )]$$

$$\Rightarrow \dfrac {dx}{d\theta }=c(-2\sin \theta +\csc \theta )$$

$$y=c\sin 2\theta $$

$$\dfrac{dy}{d\theta }=2c\cos 2\theta $$

Also, $$y_1=c\sin \dfrac{2\pi }{3}=\dfrac{c\sqrt 3}{2}$$

Slope of tangent at $$\theta =\dfrac{\pi }{3}$$ is

$$m=\dfrac{dy}{dx}_{\theta =\frac{\pi }{3}}=\sqrt 3$$

Length of tangent $$=\left | \dfrac{y_1\sqrt{1+m^2}}{m} \right |=c$$

Length of subtangent $$=\left | \dfrac{y_1}{m} \right |=\dfrac{c}{2}$$

Required sum $$=c+\dfrac{c}{2}=\dfrac{3c}{2}$$
Question 7
1 / -0

Length of the normal to the curve at any point on the curve $$y=\dfrac { a\left( { e }^{ x/a }+{ e }^{ -x/a } \right) }{ 2 } $$ varies as

A
$$x$$

B
$${ x }^{ 2 }$$

C
$$y$$

D
$${ y }^{ 2 }$$

Solution

Given,

$$y=\dfrac{a(e^{\frac{x}{a}}+e^{-\frac{x}{a}})}{2}$$

$$\Rightarrow y=a\cos h\left ( \dfrac{x}{a} \right )$$

$$\dfrac{dy}{dx}=a \sin h\left ( \dfrac{x}{a} \right )\dfrac{1}{a}$$

$$=\sin h\left ( \dfrac{x}{a} \right )$$

Length of normal $$=y\left [ 1+\left (\dfrac{dy}{dx} \right )^2 \right ]^{\frac{1}{2}}$$

$$=a\cos h\left ( \dfrac{x}{a} \right )\left [ 1+ \sin ^2h\left ( \dfrac{x}{a} \right )\right ]^{\frac{1}{2}}$$

$$=a\cos h\left ( \dfrac{x}{a} \right )\left [ \cos ^2h\left ( \dfrac{x}{a} \right ) \right ]^{\frac{1}{2}}$$

$$=a\cos ^2h\left ( \dfrac{x}{a} \right )$$

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

$$=\dfrac{y^2}{a}$$
Question 8
1 / -0

If the tangent to the curve $$x=at^2, y=2at$$ is perpendicular to $$x$$-axis, then its point of contact is

A
$$(a, a)$$

B
$$(0, a)$$

C
$$(0, 0)$$

D
$$(a, 0)$$

Solution

Let the required point be $$(x_1,y_1).$$

It is given that points lies on the curve.

$$\therefore$$ $$x_1=at^2$$ and $$y_1=2at$$

Now,
$$x=at^2$$ and $$y=2at$$

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

$$\Rightarrow$$ $$\dfrac{dx}{dt}=2at$$ and $$\dfrac{dy}{dt}=2a$$

$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$

$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{2a}{2at}$$

$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{1}{t}$$

$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{2a}{y}$$ [ Since, $$y=2at$$ ]

Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{2a}{y_1}$$

It is given that the tangent is perpendicular to the $$y-axis.$$

It means theta it is parallel to the $$x-axis.$$

$$\therefore$$ Slope of the tangent $$=$$ Slope of the $$x-axis.$$

$$\Rightarrow$$ $$\dfrac{2a}{y_1}=0$$

$$\Rightarrow$$ $$a=0$$

Now,

$$x_1=at^2=(0)t^2=0$$

$$y_1=2at=2(0)t=0$$

$$\therefore$$ $$(x_1,y_1)=(0,0)$$
Question 9
1 / -0

The slope of the tangent to the curve $$x=t^2+3t-8, y=2t^2-2t-5$$ at point $$(2, -1)$$ is

A
$$\dfrac {22}{7}$$

B
$$\dfrac {6}{7}$$

C
$$-6$$

D
$$\dfrac {7}{6}$$

Solution

$$x=t^2+3t-8$$ ........$$(1)$$
$$y=2t^2-2t-5$$ ..........$$(2)$$

Differentiate $$(1)$$, we get
$$\dfrac{dx}{dt}=2t+3$$

Differentiate $$(2)$$, we get
$$\dfrac{dy}{dx}=4t-2$$
$$m=\dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$$

Given point is $$(2, -1)$$
Put the point in original x and y, we get
$$\Rightarrow x=t^2+3t-8$$
$$2=t^2+3t-8$$
$$t^2+3t-10=0$$
$$(t-2)(t+5)=0$$
$$t=2, t=-5$$

$$\Rightarrow y=2t^2-2t-5$$
$$(-1)=2t^2-2t-5$$
$$2t^2-2t-4=0$$
$$(t+1)(t-2)=0$$
$$t=-1, t=2$$

Since, $$t=2$$ common in both parts, so we take
$$\dfrac{dy}{dx}=\dfrac{4t-2}{2t-3}$$ at $$t=2$$

At $$t=2$$
$$\dfrac{dy}{dx}=\dfrac{4(2)-2}{2(2)-3}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$$

$$m=\dfrac{dy}{dx}=\dfrac{6}{7}$$.
Question 10
1 / -0

At what points the slope of the tangent to the curve $$x^2+y^2-2x-3=0$$ is zero?

A
$$(3, 0), (-1, 0)$$

B
$$(3, 0), (1, 2)$$

C
$$(-1, 0), (1, 2)$$

D
$$(1, 2), (1, -2)$$

Solution

$$x^2+y^2-2x-3=0$$ is zero.

Differentiate w.r.t. x
$$2x+2y\dfrac{dy}{dx}-2=0$$

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

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

$$\dfrac{dy}{dx}=\dfrac{1-x}{y}$$ ........$$(1)$$

If line is parallel to x-axis
Angle with x-axis $$=\theta =0$$
Slope of x-axis$$=\tan\theta =\tan 0^o=0$$
Slope of tangent $$=$$ Slope of x-axis

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

$$\dfrac{1-x}{y}=0$$

$$x=1$$

Finding y when $$x=1$$
$$x^2+y^2-2x-3=0\\$$
$$(1)^2+y^2-2(1)-3=0\\$$
$$1+y^2-2-3=0\\$$
$$y=\pm 2$$

Hence, the points are $$(1, 2)$$ and $$(1, -2)$$.

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 41

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Tangents and its Equations Test 41

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SHARING IS CARING

Question 1

$$\dfrac { \pi }{ 2 } $$

$$\dfrac { \pi }{ 4 } $$

$$\pi +\dfrac { t }{ 2 } $$

$$\dfrac { \pi }{ 4 } +\dfrac { t }{ 2 } $$

Solution

Question 2

$$\left| \tan { \theta } \right| >1$$

$$\left| \tan { \theta } \right| <1$$

$$\left| \tan { \theta } \right| \ge1$$

$$\left| \tan { \theta } \right| \le 1$$

Solution

Question 3

$$x+y=1$$

$$x-y+1=0$$

$$x+y=2$$

$$2x-y+1=0$$

Solution

Question 4

$$1/2$$

$$2$$

$$\sqrt{2}$$

$$\cfrac { 1 }{ \sqrt { 2 } } $$

Solution

Question 5

x+y=1

x-y+1=0

2x+y=2

2x-y+1

Solution

Question 6

$$\dfrac { c }{ 2 } $$

$$2c$$

$$\dfrac { 3c }{ 2 } $$

$$\dfrac { 5c }{ 2 } $$

Solution

Question 7

$$x$$

$${ x }^{ 2 }$$

$$y$$

$${ y }^{ 2 }$$

Solution

Question 8

$$(a, a)$$

$$(0, a)$$

$$(0, 0)$$

$$(a, 0)$$

Solution

Question 9

$$\dfrac {22}{7}$$

$$\dfrac {6}{7}$$

$$-6$$

$$\dfrac {7}{6}$$

Solution

Question 10

$$(3, 0), (-1, 0)$$

$$(3, 0), (1, 2)$$

$$(-1, 0), (1, 2)$$

$$(1, 2), (1, -2)$$

Solution

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