Tangents and its Equations Test 31

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Tangents and its Equations Test 31

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Weekly Quiz Competition

Question 1
1 / -0

A normal to parabola, whose inclination is $$30^o$$, cuts it again at an angle of.

A
$$\tan^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right)$$

B
$$\tan^{-1}\left(\displaystyle\frac{2}{\sqrt{3}}\right)$$

C
$$\displaystyle\tan^{-1}\cdot(2\sqrt{3})$$

D
$$\displaystyle\tan^{-1}\left(\displaystyle\frac{1}{2\sqrt{3}}\right)$$

Solution

Equation of normal in terms of parameter
To parabola $$y^2=4ax$$ at $$(at^2, 2at)$$, the equation of normal is $$y=-tx+2at+at^3$$
To parabola $$x^2=4ay$$ at $$(2at, at^2)$$ the equation of normal is $$x+ty=2at+at^3$$

Let the normal $$P(at_1,2at_1)$$ by $$y= -t_1x + 2at_1+at_1^3$$
$$\theta$$ $$=-t_1=$$ slope of normal
It meets the curve at $$\theta$$ say $$(at_2^2,2at_2)$$
Now angle between the normal and parabola $$=$$ Angle the normal and tangent at $$\theta$$ (i.e) $$t_2= x + at_2^2$$

$$\tan \phi$$ =$$\dfrac{m_1-m_2}{1+m_1m_2}$$

$$=\dfrac{-t_1-\dfrac{1}{t_2}}{1+\dfrac{(-t_1)}{(-t_2)}}$$

$$=\dfrac{-(t_1t_2+1)}{t_2-t_1}$$

$$=\dfrac{-(t_1^2+1)}{t_2-t_1}$$

$$=\dfrac{-(-t_1^2-1)}{\dfrac{-2(1+t_1^2)}{(t_1)}}$$
$$=\dfrac{-t_1}{2}$$

$${\tan}\phi=\dfrac{\tan\theta}{2}$$
hence,
$$\phi$$=$$\tan^{-1}\dfrac{\tan30^.}{2} = \tan^{-1}(\dfrac{1}{2\times3^{\tfrac{1}{2}}})$$
Question 2
1 / -0

If the slope of the tangent to the curve $$y=a{ x }^{ 3 }+bx+4$$ at $$(2,14) = 21$$, then the values of $$a$$ and $$b$$ are respectively

A
$$2,-3$$

B
$$3,-2$$

C
$$-3,-2$$

D
$$2,3$$

Solution

$$(2,14)$$ lies on the curve $$y=ax^3+bx+4$$
=> $$8a+2b+4=14$$
=> $$4a+b=5$$----(1)

$$\dfrac{dy}{dx}=3ax^2+b|_{x=2}=21$$
=> $$12a+b=21$$--(2)

Solving equations (1) and (2), we get $$a=2, b=-3$$
Question 3
1 / -0

The angle between the curves $$x^{2} + y^{2} = 25$$ and $$x^{2} + y^{2} - 2x + 3y - 43 = 0$$ at $$(-3, 4)$$ is

A
$$\tan^{-1}(1)$$

B
$$\tan^{-1}\left (\dfrac {1}{68}\right )$$

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

D
$$\tan^{-1}\left (\dfrac {3}{4}\right )$$

Solution

slope of first curve at $$(-3,4)$$
$$m_1 = dy/dx = \dfrac{-x}{y} = \dfrac{3}{4}$$

slope of second curve at $$(-3,4)$$
$$m_2 = \dfrac{-2x+2}{2y+3} = \dfrac{8}{11}$$

angle between them $$= tan^{-1} \dfrac{m_1 - m_2}{1+m_1 +m_2}$$
$$=tan^{-1} \dfrac{1/44}{17/11} =tan^{-1} \dfrac{1}{68}$$
Question 4
1 / -0

If the tangent at each point of the curve $$y=\cfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$$ makes an acute angle with positive direction of X-axis then

A
$$a\ge 1$$

B
$$-1\le a\le 1$$

C
$$a\le -1$$

D
None of these

Solution

On differentiating, We get
$$\dfrac{dy}{dx}=2x^2-4ax+2$$
If Tangent makes acute angle with +ve x axis, then
$$\dfrac{dy}{dx}\ge0$$ For all x
$$x^2-2ax+1\ge0\\ \Rightarrow 4a^2-4\le0\Rightarrow a^2\le1\\ -1\le a\le1$$
Question 5
1 / -0

The equation of the tangent to the curve $$y=\sqrt { 9-2{ x }^{ 2 } } $$ as the point where the ordinate and the abscissa are equal , is

A
$$2x+y-\sqrt { 3 } =0$$

B
$$2x+y-3=0$$

C
$$2x-y-3\sqrt { 3 } =0$$

D
$$2x+y-3\sqrt { 3 } =0$$

Solution

Given that $${ x }_{ 1 }={ y }_{ 1 }$$
Therefore, $${ x }_{ 1 }=\sqrt { 9-2{ x }_{ 1 }^{ 2 } } $$
$$\Rightarrow { x }_{ 1 }^{ 2 }=9-2{ x }_{ 1 }^{ 2 }\quad $$
$$\Rightarrow { x }_{ 1 }=\pm \sqrt { 3 } $$
since, $${ y }_{ 1 }>0$$, therefore the point is $$\left( \sqrt { 3 } ,\sqrt { 3 } \right) $$
Also, $$y=\sqrt { 9-2{ x }^{ 2 } } $$
$$\Rightarrow { y }^{ 2 }=9-2{ x }^{ 2 }$$
On differentiating w.r.t $$x$$, we get
$$2y\cfrac { dy }{ dx } =-4x\quad $$
$$\Rightarrow \cfrac { dy }{ dx } =-\cfrac { 2x }{ y } $$
$$\Rightarrow { \left( \cfrac { dy }{ dx } \right) }_{ \left( \sqrt { 3 } ,\sqrt { 3 } \right) }=-2$$
So, the required equation of tangent is
$$\left( y-\sqrt { 3 } \right) =-2\left( x-\sqrt { 3 } \right) $$
$$\Rightarrow 2x+y-3\sqrt { 3 } =0\quad $$
Question 6
1 / -0

The equation of the curve satisfying the differential equation $$y_{2}(x^{2} + 1) = 2xy_{1}$$ passing through the point $$(0, 1)$$ and having slope of tangent at $$x = 0$$ as $$3$$ is

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

B
$$y = x^{2} + 3x + 1$$

C
$$y = x^{3} + 3x + 1$$

D
None of these

Solution

Given $$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \left( { x }^{ 2 }+1 \right) =2x\cfrac { dy }{ dx } $$
$$y={ x }^{ 2 }+3x+1$$
Slope, $${ \left( \cfrac { dy }{ dx } \right) }_{ x=0 }={ (2x+3) }_{ x=0 }\Rightarrow 2\times 0+3\Rightarrow 3$$
At point $$x=0$$
$$y={ (0) }^{ 2 }+3\times 0+1$$
Thus this curve passes through point $$(0,1)$$ and have slope of tangent at $$x=0$$ as $$3$$
$$y={ x }^{ 2 }+3x+1$$ is the answer
Question 7
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The slope of the tangent at each point of the curve is equal to the sum of the coordinate of the point. Then, the curve that passes through the origin is

A
$$x + y = e^{x} - 1$$

B
$$e^{x} = x + y$$

C
$$y = e^{x}$$

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

Solution

Given, $$\dfrac {dy}{dx}=x+y\Rightarrow dy=xdx+ydx\Rightarrow dy+dx=xdx+ydx+dx$$

So, $$d(x+y)=(x+y+1)dx\Rightarrow dx=\dfrac {d(x+y+1)}{x+y+1}$$
Integrating on both sides,
$$x=\ln(x+y+1)$$
On taking antilog,
$$e^x=x+y+1$$
Or. $$e^x-1=x+y$$
So, Option A is correct answer.
Question 8
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The equation of tangent to the curve $${ \left( \cfrac { x }{ a } \right) }^{ n }+{ \left( \cfrac { y }{ b } \right) }^{ n }=2\quad $$ at the point $$(a,b)$$ is

A
$$\cfrac { x }{ a } =-\cfrac { y }{ b } $$

B
$$\cfrac { x }{ a } +\cfrac { y }{ b } =2$$

C
$$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =n$$

D
$$\cfrac { x }{ a } =\cfrac { y }{ b } $$

Solution

Given equation of the curve, $$\left( \dfrac{x}{a} \right) ^n + \left( \dfrac{y}{b} \right) ^n =2$$

Differente the above equation to find the slope of the curve

$$ n{ \left( \dfrac { x }{ a } \right) }^{ n-1 }.\dfrac { 1 }{ a } +n{ \left( \dfrac { y }{ b } \right) }^{ n-1 }.\dfrac { 1 }{ b } \dfrac { dy }{ dx } =0$$

$$ \implies \dfrac { dy }{ dx } =\dfrac { -b }{ a } { \left( \dfrac { x }{ a } \right) }^{ n-1 }{ \left( \dfrac { b }{ y } \right) }^{ n-1 }$$

$$ \therefore \,\,$$Slope at $$(a,b) = \dfrac { -b }{ a } { \left( \dfrac { a }{ a } \right) }^{ n-1 }{ \left( \dfrac { b }{ b } \right) }^{ n-1 } = \dfrac{-b}{a}$$

$$\therefore \,\,$$Equation of the tangent at $$(a,b)$$ is:

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

$$ bx +ay = 2ab$$

$$ \dfrac{x}{a}+\dfrac{y}{b} = 2$$
Question 9
1 / -0

The slope of the tangent at the point $$(h, h)$$ of the circle $$x^{2} + y^{2} = a^{2}$$ is :

A
$$0$$

B
$$1$$

C
$$-1$$

D
Depends on $$h$$

Solution

Slope of the tangent for the cicle $$x^{2}+y^{2}=a^{2}$$ can be calculated by differentiating the circle's equation.
On differentiating w.r.t x, $$2x + 2y\dfrac{dy}{dx}=0$$
Slope $$m$$$$=\space \dfrac{dy}{dx}=-\dfrac{x}{y}$$
Point given is $$(h,h)$$
Putting point in the slope equation $$m=\dfrac{dy}{dx}=-\dfrac{h}{h}$$
Slope $$m=-1$$
Hence, $$(C)$$
Question 10
1 / -0

The angle at which the curve $$y={ x }^{ 2 }$$ and the curve $$x=\cfrac { 5 }{ 3 } \cos { t } ,y=\cfrac { 5 }{ 4 } \sin { t } $$ intersect is

A
$$\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

B
$$\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

C
$$-\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

D
$$2\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

Solution

Given
$$y={ x }^{ 2 }...(i)$$
$$\quad x=\cfrac { 5 }{ 3 } \cos { t } ;y=\cfrac { 5 }{ 4 } \sin { t } ...(ii)$$
Which is parametric equation, we change this equation is caresian equation as follows
$$\cos { t } =\cfrac { 3 }{ 5 } x;\sin { t } =\cfrac { 4 }{ 5 } y$$
On Squaring and adding both i.e., $$\cos{t}$$ and $$\sin {t}$$ we get
$$\cfrac { 9 }{ 25 } { x }^{ 2 }+\cfrac { 16 }{ 25 } { y }^{ 2 }=\cos ^{ 2 }{ t } +\sin ^{ 2 }{ t } $$
$$\Rightarrow 9{ x }^{ 2 }+16{ y }^{ 2 }=25....(iii)\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right] $$
$$\therefore$$ The intersection points at Eq.(i) and (iii) are $$(1,1)$$ and $$(-1,1)$$
Now, slope of tangent of Eq.(i) at point $$(1,1)$$ is
$${ m }_{ 1 }=\cfrac { dy }{ dx } =2x\quad \therefore { m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (1,1) }=2$$
And slope of tangent of Eq. (iii) at point $$(1,1)$$ is
$${ m }_{ 2 }=\cfrac { dy }{ dx } =-\cfrac { 9 }{ 16 } $$
$$\therefore$$ Angle at point of intersection of Eqs. (i) and (iii) we get
$${ \theta }_{ 1 }=\tan ^{ -1 }{ \left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$
similarly, slope of tangent of Eq. (i) at point $$(-1,1)$$
$${ m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (-1,1) }=-2$$
And slope of tangent of Eq. (iii) at point $$(-1,1)$$
$${ m }_{ 2 }=\cfrac { dy }{ dx } =\cfrac { 9 }{ 16 } $$
$$\therefore$$ Angle at point of intersection of Eqs. (i) and (iii) we get
$${ \theta }_{ 2 }=\tan ^{ -1 }{ \left| \cfrac {-2- \cfrac { 9 }{ 16 } }{ 1-\cfrac { 18 }{ 16 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } \quad \quad $$

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Tangents and its Equations Test 31

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SHARING IS CARING

Question 1

$$\tan^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right)$$

$$\tan^{-1}\left(\displaystyle\frac{2}{\sqrt{3}}\right)$$

$$\displaystyle\tan^{-1}\cdot(2\sqrt{3})$$

$$\displaystyle\tan^{-1}\left(\displaystyle\frac{1}{2\sqrt{3}}\right)$$

Solution

Question 2

$$2,-3$$

$$3,-2$$

$$-3,-2$$

$$2,3$$

Solution

Question 3

$$\tan^{-1}(1)$$

$$\tan^{-1}\left (\dfrac {1}{68}\right )$$

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

$$\tan^{-1}\left (\dfrac {3}{4}\right )$$

Solution

Question 4

$$a\ge 1$$

$$-1\le a\le 1$$

$$a\le -1$$

None of these

Solution

Question 5

$$2x+y-\sqrt { 3 } =0$$

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

$$2x-y-3\sqrt { 3 } =0$$

$$2x+y-3\sqrt { 3 } =0$$

Solution

Question 6

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

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

$$y = x^{3} + 3x + 1$$

None of these

Solution

Question 7

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

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

$$y = e^{x}$$

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

Solution

Question 8

$$\cfrac { x }{ a } =-\cfrac { y }{ b } $$

$$\cfrac { x }{ a } +\cfrac { y }{ b } =2$$

$$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =n$$

$$\cfrac { x }{ a } =\cfrac { y }{ b } $$

Solution

Question 9

$$0$$

$$1$$

$$-1$$

Depends on $$h$$

Solution

Question 10

$$\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

$$\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

$$-\tan ^{ -1 }{ \cfrac { 2 }{ 41 } } $$

$$2\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } $$

Solution

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