Tangents and its Equations Test 28

Result

Tangents and its Equations Test 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

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.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 28

Result

Tangents and its Equations Test 28

Score

-

Rank

-

SHARING IS CARING

Question 1

1

2

3

No tangent is parallel to x-axis.

Solution

Question 2

000

−1-1−1

111

Not defined

Solution

Question 3

xa+yb=2\dfrac {x}{a} + \dfrac {y}{b} = 2ax​+by​=2

xa+yb=12\dfrac {x}{a} + \dfrac {y}{b} = \dfrac {1}{2}ax​+by​=21​

xb−ya=2\dfrac {x}{b} - \dfrac {y}{a} = 2bx​−ay​=2

ax+by=2ax + by = 2ax+by=2

Solution

Question 4

0

1

2

None of the above

Solution

Question 5

∞\infty∞

111111

919\displaystyle\frac{9}{19}199​

−199\displaystyle-\frac{19}{9}−919​

Solution

Question 6

138\dfrac{13}{8}813​

74\dfrac{7}{4}47​

158\dfrac{15}{8}815​

222

Solution

Question 7

ax+by=1ax+by=1ax+by=1

xa−yb=1\dfrac { x }{ a } -\dfrac { y }{ b } =1ax​−by​=1

xa+yb=1\dfrac { x }{ a } + \dfrac { y }{ b } =1ax​+by​=1

ax−by=1ax-by=1ax−by=1

Solution

Question 8

(2,−1)(2, -1)(2,−1)

(10,3)(10, 3)(10,3)

(2,1)(2, 1)(2,1)

(5,−2)(5, -2)(5,−2)

Solution

Question 9

(1,0)(1, 0)(1,0)

(2,0)(2, 0)(2,0)

(−12,0)\left(-\dfrac{1}{2}, 0\right)(−21​,0)

(12,0)\left(\dfrac{1}{2}, 0\right)(21​,0)

Solution

Question 10

2x−3y=12x - 3y = 12x−3y=1

x=0x = 0x=0

x=1x = 1x=1

y=0y = 0y=0

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$0$

$-1$

$1$

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

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

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

$ax + by = 2$

$\infty$

$11$

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

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

$\dfrac{13}{8}$

$\dfrac{7}{4}$

$\dfrac{15}{8}$

$2$

$ax+by=1$

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

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

$ax-by=1$

$(2, -1)$

$(10, 3)$

$(2, 1)$

$(5, -2)$

$(1, 0)$

$(2, 0)$

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

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

$2x - 3y = 1$

$x = 0$

$x = 1$

$y = 0$