Tangents and its Equations Test 55

Result

Tangents and its Equations Test 55

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

Paraboals $(y-\alpha )^{2}=4a(x-\beta )and (y-\alpha )^{2}=4a'(x-\beta ')$ will have a common normal (other than the normal passing through vertex ofparabola)if:

A
$\frac{2(a-a')}{\beta '-\beta }< 1$

B
$\frac{2(a-a')}{\beta -\beta' }< 1$

C
$\frac{2(a'-a)}{\beta -\beta' }< 1$

D
$\frac{2(a'a)}{\beta -\beta' }> 1$

Solution

Login to see Solution & AIR
Question 2
1 / -0

The normal to the curve $x=a\left( \cos { \theta } +\theta \sin { \theta } \right)$ , $y=a\left( \sin { \theta } -\theta \cos { \theta } \right)$ at any point $\theta$ is such that

A
it passes through origin

B
it passes through the point $(1,1)$

C
it passes through $\left( \frac { a\pi }{ 2 } ,-a \right)$

D
it is at a constant distance from the origin

Solution

$x=a\cos \theta +a\theta \sin \theta$
$\dfrac {dx}{d\theta }=-a\sin \theta +a\left\{\theta (-\cos \theta )+\sin \theta \right\}$
$=-a\theta \cos \theta$
$y=a(\sin \theta -\theta \cos \theta )$
$\dfrac {dy}{dx}=a\cos \theta -a \left\{\theta \sin \theta +\cos \theta \right\}$
$\therefore \ \dfrac {dy}{dx}=\dfrac {-a\theta \sin \theta }{-a\theta \cos \theta }=\tan \theta$
The equation of normal at $\theta$ is:
$y-y_1=-\dfrac {1}{\dfrac {dy}{dx}} (x-x_1)$
or, $y-a (\sin \theta -\theta \cos \theta )=-\dfrac {\cos \theta }{\sin \theta }(x-a\cos \theta -a\theta \sin \theta )$
or, $y \sin \theta -\sin^2\theta +a\theta \sin \theta \cos \theta$
$=-x\cos \theta +a\cos^2\theta +a\theta \sin \theta \cos \theta$
or, $y\sin \theta +x\cos \theta =a$
which wakes it clear that normal passes through origin $(0,0)$

Login to see Solution & AIR
Question 3
1 / -0

The slope of the curve $y=\sin { x } +\cos ^{ 2 }{ x }$ is zero at a point , whose x-coordinate can be ?

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

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

C
${ \pi }$

D
$\dfrac { \pi }{ 3 }$

Solution

Login to see Solution & AIR
Question 4
1 / -0

The tangent to the circle $x^2+y^2=5$ at the point $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$ at

A
$(-2,1)$

B
$(-3,0)$

C
$(-1,-1)$

D
$(3,-1)$

Solution

Login to see Solution & AIR
Question 5
1 / -0

If for a curve represented parametrically by $x={ sec }^{ 2 }t,\quad y=cot\quad t\quad$ , the tangent at a point $P(t=\frac { \pi }{ 4 } )$ meets the curve again at the point Q, then $\begin{vmatrix} PQ \end{vmatrix}$ is equal to

A
$\frac { 2\sqrt { 5 } }{ 3 }$

B
$\frac { 3\sqrt { 5 } }{ 2 }$

C
$\frac { 5\sqrt { 3 } }{ 3 }$

D
$\frac { 5\sqrt { 5 } }{ 4 }$

Solution

Login to see Solution & AIR
Question 6
1 / -0

If the subnormal to the curve ${ x }^{ 2 }.{ y }^{ n }={ a }^{ 2 }$ is a constant then n=

A
$-4$

B
$-3$

C
$-2$

D
$-1$

Solution

Given,

$x^2y^n=a^2$

$\Rightarrow x=\sqrt{\dfrac{a^2}{y^n}}$

$\Rightarrow 2xy^n+x^2ny^{n-1}\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{nx}$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2y}{n\sqrt{\frac{a^2}{y^n}}}$

$\therefore \dfrac{dy}{dx}=-\dfrac{2}{na}y^{\frac{n+2}{2}}$

Subnormal is given by,

$SN=y \dfrac{dy}{dx}$

$=y\left ( -\dfrac{2}{na}y^{\frac{n+2}{2}} \right )$

$\therefore SN=-\dfrac{2}{na}y^{\frac{n+4}{2}}$

$\therefore n=-4$

Login to see Solution & AIR
Question 7
1 / -0

If $\theta$ is angle of intersection between $y=10-x^{2}$ and $y=4+x^{2}$ then $|\tan \theta|$ is-

A
$\dfrac {5\sqrt {3}}{11}$

B
$\dfrac {7\sqrt {3}}{15}$

C
$\dfrac {4\sqrt {3}}{11}$

D
$none$

Solution

Given: $y=10-{x}^{2}$ ...... $(1)$
$y=4+{x}^{2}$ ...... $(2)$
On solving,
$10-{x}^{2}=4+{x}^{2}$
$\Rightarrow 2{x}^{2}=6$
$\Rightarrow {x}^{2}=\dfrac{6}{2}=3$
$\therefore x=\pm \sqrt{3}$ from $(1)$
$\Rightarrow y=10-{\left(\pm\sqrt{3}\right)}^{2}=10-3=7$
Consider $y=10-{x}^{2}$
$\dfrac{dy}{dx}=-2x$
${m}_{1}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=-2\times\sqrt{3}=-2\sqrt{3}$
Consider $y=4+{x}^{2}$
$\dfrac{dy}{dx}=2x$
${m}_{2}=\left[\dfrac{dy}{dx}\right]_{\left(\sqrt{3},7\right)}=2\times\sqrt{3}=2\sqrt{3}$
$=\left|\dfrac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}\right|$
$\tan{\theta}=\left|\dfrac{-2\sqrt{3}-2\sqrt{3}}{1+\left(-2\sqrt{3}\right)\left(2\sqrt{3}\right)}\right|$
$=\left|\dfrac{-4\sqrt{3}}{1-12}\right|$
$=\dfrac{4\sqrt{3}}{11}$
$\therefore \tan{\theta}=\dfrac{4\sqrt{3}}{11}$

Login to see Solution & AIR
Question 8
1 / -0

$y = 6\tan \,x\left( {{e^x} - x - 1} \right) - 3{x^3} - {x^4} - \frac{5}{4}{x^5},\,$ if ${n^{th}}$ derivative at x=0 is non zero then least value of n is

A
3

B
4

C
5

D
6

Solution

Login to see Solution & AIR
Question 9
1 / -0

The curve $y-{e}^{xy}+x=0$ has a vertical tangent at the point

A
$(1,1)$

B
At no point

C
$(0,1)$

D
$(1,0)$

Solution

Login to see Solution & AIR
Question 10
1 / -0

The slope of the tangent to the curve at a point (x, y) on it is proportional to (x-2). If the slope of the tangent to the curve at (10,-9) on it -3. The equation of the curve is

A
$y=k{ \left( x-2 \right) }^{ 2 }$

B
$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +1$

C
$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +3$

D
$y=k{ \left( x+2 \right) }^{ 2 }$

Solution

${\textbf{Step -1: Find the value of slope}}{\textbf{.}}$
${\text{The slope of the tangent to the curve at a point }}\left( {x,y} \right){\text{ on it is proportional to }}\left( {x - 2} \right).$
$= \dfrac{{dy}}{{dx}} = m\left( {x - 2} \right)$
$= \dfrac{{dy}}{{dx}} = k\left( {x - 2} \right)\left[ {{\text{where }}k{\text{ is a proportionality constant}}} \right]$
${\text{Slope of the tangent to the curve at }}\left( {10, - 9} \right){\text{ is }} - 3.$
$\Rightarrow - 3 = k\left( {10 - 2} \right)$
$\Rightarrow - 3 = 8k$
$\Rightarrow k = - \dfrac{3}{8}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{3}{8}\left( {x - 2} \right)$
${\textbf{Step -2: Integrate the formed equation and solve it further}}{\textbf{.}}$
${\text{Integrating both sides,}}$
$\Rightarrow y = - \dfrac{3}{8}\dfrac{{{{\left( {x - 2} \right)}^2}}}{2} + c{\text{ }}.......\left( i \right){\text{ }}\left[ {{\text{Eqaution of curve}}} \right]$
${\text{Also, }}\left( {10, - 9} \right){\text{ satisfy the equation of curve as it lies on the curve}}{\text{.}}$
$\therefore - 9 = - \dfrac{3}{8}\dfrac{{{{\left( {10 - 2} \right)}^2}}}{2} + c$
$\Rightarrow - 9 = - \dfrac{3}{{16}} \times {8^2} + c$
$\Rightarrow c = 3$
${\textbf{Hence, the equation of the curve is }}\mathbf{y= - \dfrac{3}{{16}}{\left( {x - 2} \right)^2} + 3.}$

Login to see Solution & AIR

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 55

Result

Tangents and its Equations Test 55

Score

-

Rank

-

SHARING IS CARING

Question 1

2(a−a′)β′−β<1\frac{2(a-a')}{\beta '-\beta }< 1β′−β2(a−a′)​<1

2(a−a′)β−β′<1\frac{2(a-a')}{\beta -\beta' }< 1β−β′2(a−a′)​<1

2(a′−a)β−β′<1\frac{2(a'-a)}{\beta -\beta' }< 1β−β′2(a′−a)​<1

2(a′a)β−β′>1\frac{2(a'a)}{\beta -\beta' }> 1β−β′2(a′a)​>1

Solution

Question 2

it passes through origin

it passes through the point (1,1)(1,1)(1,1)

it passes through (aπ2,−a)\left( \frac { a\pi }{ 2 } ,-a \right) (2aπ​,−a)

it is at a constant distance from the origin

Solution

Question 3

π4\dfrac { \pi }{ 4 } 4π​

π2\dfrac { \pi }{ 2 } 2π​

π{ \pi }π

π3\dfrac { \pi }{ 3 } 3π​

Solution

Question 4

(−2,1)(-2,1)(−2,1)

(−3,0)(-3,0)(−3,0)

(−1,−1)(-1,-1)(−1,−1)

(3,−1)(3,-1)(3,−1)

Solution

Question 5

253\frac { 2\sqrt { 5 } }{ 3 } 325​​

352\frac { 3\sqrt { 5 } }{ 2 } 235​​

533\frac { 5\sqrt { 3 } }{ 3 } 353​​

554\frac { 5\sqrt { 5 } }{ 4 } 455​​

Solution

Question 6

−4-4−4

−3-3−3

−2-2−2

−1-1−1

Solution

Question 7

5311\dfrac {5\sqrt {3}}{11}1153​​

7315\dfrac {7\sqrt {3}}{15}1573​​

4311\dfrac {4\sqrt {3}}{11}1143​​

nonenonenone

Solution

Question 8

3

4

5

6

Solution

Question 9

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

At no point

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

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

Solution

Question 10

y=k(x−2)2y=k{ \left( x-2 \right) }^{ 2 }y=k(x−2)2

y=−316(x−2)2 +1y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +1y=16−3​(x−2)2 +1

y=−316(x−2)2 +3y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +3y=16−3​(x−2)2 +3

y=k(x+2)2y=k{ \left( x+2 \right) }^{ 2 }y=k(x+2)2

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.

$\frac{2(a-a')}{\beta '-\beta }< 1$

$\frac{2(a-a')}{\beta -\beta' }< 1$

$\frac{2(a'-a)}{\beta -\beta' }< 1$

$\frac{2(a'a)}{\beta -\beta' }> 1$

it passes through the point $(1,1)$

it passes through $\left( \frac { a\pi }{ 2 } ,-a \right)$

$\dfrac { \pi }{ 4 }$

$\dfrac { \pi }{ 2 }$

${ \pi }$

$\dfrac { \pi }{ 3 }$

$(-2,1)$

$(-3,0)$

$(-1,-1)$

$(3,-1)$

$\frac { 2\sqrt { 5 } }{ 3 }$

$\frac { 3\sqrt { 5 } }{ 2 }$

$\frac { 5\sqrt { 3 } }{ 3 }$

$\frac { 5\sqrt { 5 } }{ 4 }$

$-4$

$-3$

$-2$

$-1$

$\dfrac {5\sqrt {3}}{11}$

$\dfrac {7\sqrt {3}}{15}$

$\dfrac {4\sqrt {3}}{11}$

$none$

$(1,1)$

$(0,1)$

$(1,0)$

$y=k{ \left( x-2 \right) }^{ 2 }$

$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +1$

$y=\frac { -3 }{ 16 } { \left( x-2 \right) }^{ 2 }\ +3$

$y=k{ \left( x+2 \right) }^{ 2 }$