Tangents and its Equations Test 28

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Tangents and its Equations Test 28

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Question 1
1 / -0

How many tangents are parallel to x-axis for the curve $$ y = x^2 - 4x + 3$$ ?

A
1

B
2

C
3

D
No tangent is parallel to x-axis.

Solution

Slope of the tangent is $$\dfrac {dy} {dx}$$

So, $$\dfrac{dy}{dx}=2x-4$$

Tangent parallel to x-axis. So, slope of tangent should be $$0.$$

$$\Rightarrow 2x-4=0$$

$$\Rightarrow x=2$$ is the only point where slope is parallel to x-axis.

So, only 1 tangent exists.
Question 2
1 / -0

The slope of the tangent to the curve given by $$x = 1 - \cos { \theta }$$, $$y = \theta -\sin { \theta } $$ at $$\theta = \dfrac { \pi }{ 2 } $$ is

A
$$0$$

B
$$-1$$

C
$$1$$

D
Not defined

Solution

$$y=\theta -\sin\theta$$
$$\Longrightarrow \dfrac { dy }{ d\theta } =1-\cos\theta$$
$$x=1-\cos\theta$$
$$\Longrightarrow \dfrac { dx }{ d\theta } =sin\theta $$
Slope of the tangent is $$\dfrac{dy}{dx} $$
$$\Longrightarrow \dfrac{dy}{dx} =\dfrac { 1-\cos\theta}{\sin\theta}$$

At $$\theta =\dfrac { \pi }{ 2 } $$
$$\Longrightarrow \dfrac { dy }{ dx } =\dfrac { 1-\cos\frac { \pi }{ 2 } }{ \sin\frac { \pi }{ 2 } } =1$$
Question 3
1 / -0

The equations of tangent to the curve $$\left (\dfrac {x}{a}\right )^{n} + \left (\dfrac {y}{b}\right )^{n} = 2at, (a, b)$$ is $$5$$.

A
$$\dfrac {x}{a} + \dfrac {y}{b} = 2$$

B
$$\dfrac {x}{a} + \dfrac {y}{b} = \dfrac {1}{2}$$

C
$$\dfrac {x}{b} - \dfrac {y}{a} = 2$$

D
$$ax + by = 2$$

Solution

$${ \left( \dfrac { x }{ a } \right) }^{ n }+{ \left( \dfrac { y }{ b } \right) }^{ n }=2at$$
$$n{ \left( \dfrac { x }{ a } \right) }^{ n-1 }\dfrac { 1 }{ a } +n{ \left( \dfrac { y }{ b } \right) }^{ n-1 }\dfrac { y\prime }{ b } =0$$
so, $$\dfrac { y\prime }{ b } +\dfrac { 1 }{ a } =0$$ at $$(a,b)$$
so, $$y\prime =\dfrac { -b }{ a } $$
so, tangent equation
$$\left( y-b \right) =\dfrac { -b }{ a } (x-a)$$
$$ay-ab=-bx+ab$$
$$bx+ay=2ab$$
$$\boxed { \dfrac { x }{ a } +\dfrac { y }{ b } =2 } $$
Question 4
1 / -0

What is the slope of the tangent to the curve $$y=sin^{-1}(sin^2x)$$ at $$x=0$$ ?

A
0

B
1

C
2

D
None of the above

Solution

Slope of tangent to a curve f(x), at a point $$({x}_{o},{y}_{o})$$ is given by $$f'(x)$$,
where $$f'(x)$$ is derivative of f(x) at $$x={x}_{o}$$
$$f(x) =\sin^{-1}({\sin ^{ 2 }{ x } })$$
$${x}_{o}=0$$
$$f'\left( x \right) =\cfrac { df\left( x \right) }{ dx } =\cfrac { df(x) }{ d\sin ^{ 2 }{ x } } \times \cfrac { d\sin ^{ 2 }{ x } }{ d\sin x } \times \cfrac { d\sin x }{ dx } $$
$$f'\left( x \right) =\cfrac { 1 }{ \sqrt { 1-\sin ^{ 2 }{ x } } } \times 2\sin x\times \cos x$$
$$f'\left( 0 \right) =0$$
Slope of tangent to the curve is 0
Question 5
1 / -0

Find the slope of the normal to the curve $$4x^3+6x^2-5xy-8y^2+9x+14=0$$T the point $$-2, 3$$.

A
$$\infty$$

B
$$11$$

C
$$\displaystyle\frac{9}{19}$$

D
$$\displaystyle-\frac{19}{9}$$

Solution

The given curve is $$4x^3+6x^2-5xy-8y^2+9x+14=0$$

Differentiating above equation w.r.t $$x$$, we get

$$12{ x }^{ 2 }+12x-5x\cfrac { dy }{ dx } -5y-16y\cfrac { dy }{ dx } +9=0$$

$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 12x^{ 2 }+12x-5y+9 }{ \left( 5x+16y \right) } $$

At $$\left( x,y \right) \equiv \left( -2,3 \right) $$

$$m=\cfrac { dy }{ dx } =\dfrac{12(4)+12(-2)-5(3)+9}{5(-2)+16(3)}=\dfrac{18}{38}=\dfrac{9}{19}$$ is the slope of tangent to the curve
Slope of normal is $$-\dfrac{1}{m}=-\dfrac{19}{9}$$
Question 6
1 / -0

A mirror in the first quadrant is in the shape of a hyperbola whose equation is xy = 1. A light source in the second quadrant emits a beam of light that hits the mirror at the point (2,1/2). If the reflected ray is parallel to the y-axis the slope of the incident beam is

A
$$\dfrac{13}{8}$$

B
$$\dfrac{7}{4}$$

C
$$\dfrac{15}{8}$$

D
$$2$$

Solution

Slope of tangent at $$\left ( 2, \frac{1}{2} \right )$$
$$m=-\dfrac{1}{4}$$
$$tan \theta =-\dfrac{1}{y}$$

$$\theta =\phi -90^o$$
$$tan\theta =4$$
Slope of incident ray $$= m$$
$$\left | \dfrac{m-\left ( -\dfrac{1}{4} \right )}{1+m\left ( -\dfrac{1}{4} \right )} \right |=tan \theta =4$$

$$\left | \dfrac{4m+1}{4-m} \right |=4$$

$$4m + 1 = 16 - 4m$$
$$m=\dfrac{15}{8}$$
Question 7
1 / -0

The equation to the tangent to the curve $$y={ be }^{ { -x }/{ a } }$$ at the point where it crosses the $$y$$-axis is

A
$$ax+by=1$$

B
$$\dfrac { x }{ a } -\dfrac { y }{ b } =1$$

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

D
$$ax-by=1$$

Solution

Slope of the tangent to $$f$$ is $$f'$$
So, $$y'=\dfrac { d }{ dx } (b{ e }^{-x/a})=b\left(\dfrac { -1 }{ a }\right) {e }^{-x/a}$$
When it crosses $$y-axis$$, point is $$(0,b)$$.
At $$(0,b)$$ slope is $$-\dfrac{b}{a}$$ and is passes through point $$(0,b)$$
Then, the equation of line is $$y-b=-\dfrac{b}{a}(x-0)$$
$$\implies ay-ab=-bx$$
$$\implies \dfrac { x }{ a } +\dfrac { y }{ b } =1$$ is the equation of tangent.
Question 8
1 / -0

The point on the curve $$y = \sqrt {x - 1}$$ where the tangent is perpendicular to the line $$2x + y - 5 = 0$$ is

A
$$(2, -1)$$

B
$$(10, 3)$$

C
$$(2, 1)$$

D
$$(5, -2)$$

Solution

$$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {x - 1}} = m_{1}$$ is the slope of tangent to $$y=\sqrt{x-1}$$
Slope of the line $$2x + y - 5 = 0$$ is $$m_{2} = -2$$
For lines are perpendicular
$$m_{1} m_{2} = -1$$
$$\implies \left (\dfrac {1}{2\sqrt {x - 1}}\right )(-2) = -1$$
$$\implies \dfrac {2}{2\sqrt {x - 1}} = 1$$
$$\implies \sqrt {x - 1} = 1$$
Squaring both sides,
$$\implies x - 1 = 1$$
$$\implies x = 2$$
$$\therefore y = \sqrt {x - 1}$$
$$= \sqrt {2 - 1}$$
$$= \sqrt {1}$$
$$\therefore y = 1$$
$$\therefore (2, 1)$$ is the point on the curve $$y=\sqrt{x-1}$$
Question 9
1 / -0

Consider the curve $$y = e^{2x}$$.Where does the tangent to the curve at (0, 1) meet the x-axis ?

A
$$(1, 0)$$

B
$$(2, 0)$$

C
$$\left(-\dfrac{1}{2}, 0\right)$$

D
$$\left(\dfrac{1}{2}, 0\right)$$

Solution

Slope of tangent at any point to the curve is given by $$\left|{ \cfrac { df\left( x \right) }{ dx } }\right|_{ ({ x }_{ 0 }, { y }_{ 0 }) }$$
Here $$f(x)=y={e}^{2x}$$
$$\Rightarrow \cfrac{dy}{dx}=2{e}^{2x}$$
At point $$(0,1)$$
$$\cfrac{dy}{dx}=2{e}^{2(0)}$$
$$\cfrac{dy}{dx}=2$$
Equation of tangent is at point $$({x}_{0},{y}_{0})$$
$$y-{y}_{0}=m(x-{x}_{0})$$
So, tangent at (0,1) is
$$y-1=2(x-0)$$
$$\therefore y=2x+1$$
When this tangent meets x-axis, $$y$$ co-ordinate becomes zero
Putting $$y$$ in above equation as zero, we get
$$0=2x+1$$
$$\Rightarrow x=\cfrac{-1}{2}$$
Thus, the tangent to the curve meets at $$\left(-\dfrac{1}{2},0\right)$$
Question 10
1 / -0

The equation to the normal to the hyperbola $$\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$$ at $$(-4, 0)$$ is.

A
$$2x - 3y = 1$$

B
$$x = 0$$

C
$$x = 1$$

D
$$y = 0$$

Solution

$$\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$$
Differentiating above equation, we get
$$ \dfrac { { 2x } }{ 16 } -\dfrac { 2{ y } }{ 9 } .\dfrac { dy }{ dx } =0$$
$$\implies \dfrac{dy}{dx}=\dfrac{9x}{16y}$$
$$\implies m=\left|\dfrac { dy }{ dx }\right|_{(-4,0)} =\dfrac{9(-4)}{ 16(0)} =\infty$$ is the slope of tangent
Slope of the normal is $$-\dfrac{1}{m} =0$$
Slope of the normal is zero so it is parallel to $$y$$-axis.
Equation of normal is $$y-0=0(x-(-4))$$
$$\therefore y=0$$ is the equation of normal.

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Tangents and its Equations Test 28

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Question 1

1

2

3

No tangent is parallel to x-axis.

Solution

Question 2

$$0$$

$$-1$$

$$1$$

Not defined

Solution

Question 3

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

$$\dfrac {x}{a} + \dfrac {y}{b} = \dfrac {1}{2}$$

$$\dfrac {x}{b} - \dfrac {y}{a} = 2$$

$$ax + by = 2$$

Solution

Question 4

0

1

2

None of the above

Solution

Question 5

$$\infty$$

$$11$$

$$\displaystyle\frac{9}{19}$$

$$\displaystyle-\frac{19}{9}$$

Solution

Question 6

$$\dfrac{13}{8}$$

$$\dfrac{7}{4}$$

$$\dfrac{15}{8}$$

$$2$$

Solution

Question 7

$$ax+by=1$$

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

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

$$ax-by=1$$

Solution

Question 8

$$(2, -1)$$

$$(10, 3)$$

$$(2, 1)$$

$$(5, -2)$$

Solution

Question 9

$$(1, 0)$$

$$(2, 0)$$

$$\left(-\dfrac{1}{2}, 0\right)$$

$$\left(\dfrac{1}{2}, 0\right)$$

Solution

Question 10

$$2x - 3y = 1$$

$$x = 0$$

$$x = 1$$

$$y = 0$$

Solution

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