Tangents and its Equations Test 3

Result

Tangents and its Equations Test 3

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

Normal to the curve $$\displaystyle x^{2}=4y$$ which passes through the point $$(1,2)$$

A
$$x+y=3$$

B
$$x-y=3$$

C
$$2x+y=4$$

D
$$x+2y=5$$

Solution

From the equation of the curve $$\displaystyle x^{2}=4y$$,
we get

$$dy/dx=x/2.$$ If $$m$$ be the slope of the normal to $$\displaystyle x^{2}=4y$$

then $$\displaystyle

m=\dfrac{-1}{\left(\frac{dy}{dx}\right)}=-\frac{2}{x}$$

$$\displaystyle

\therefore x=\dfrac{-2}m$$ and $$y=\dfrac{x^{2}}4=1/m^{2}$$
Thus normal is

$$\displaystyle y-\frac{1}{m^{2}}=m\left ( x+\frac{2}{m} \right

)$$.

If it passes through the point (1,2),

then $$\displaystyle \therefore

2-\frac{1}{m^{2}}=m\left ( 1+\frac{2}{m} \right

)=m+2 \therefore m^{3}=-1$$

or$$\displaystyle

m=-1$$

$$ \therefore $$Required normal is,

$$(y-2)=-1(x-1) $$

$$\Rightarrow x+y=3$$
Question 2
1 / -0

Find the equation of a line passing through $$(-2,3)$$ and parallel to tangent at origin for the circle $$\displaystyle x^{2}+y^{2}+x-y=0$$

A
$$x -2 y + 5 = 0$$

B
$$x -4 y + 3 = 0$$

C
$$x - y + 5 = 0$$

D
$$2x - y + 6 = 0$$

Solution

Given,$$x^{2}+y^{2}+x-y=0$$
Differentiating w.r.t $$x$$, we get
$$2x+2y\cfrac{dy}{dx}+1-\cfrac{dy}{dx}=0$$
$$\Rightarrow \cfrac{dy}{dx}=\cfrac{1+2x}{1-2y}$$
Thus slope of the tangent at origin is, $$m=\left(\cfrac{dy}{dx}\right)_{(0,0)}=1$$
Hence required line through $$(-2,3)$$ is, $$(y-3)=1(x+2)\Rightarrow x-y+5=0$$
Question 3
1 / -0

Find the equations of tangents to parabola $$\displaystyle y^{2}= 4ax$$ which are drawn from the point (2a,3a).

A
$$\displaystyle x-y+a= 0, x-2y+4a= 0$$

B
$$\displaystyle x-y-a= 0, x-2y-4a= 0$$

C
$$\displaystyle x+y+2a= 0, x-2y+a= 0$$

D
$$\displaystyle x+y-2a= 0, x+2y-4a= 0$$

Solution

Any tangent to parabola is given by, $$\displaystyle y= mx+\frac{a}{m}$$
Now given it passes through $$(2a, 3a)$$
$$\Rightarrow \displaystyle 3a= 2am+\dfrac{a}{m}\Rightarrow 2m^2-3m+1=0\Rightarrow m =1,\dfrac{1}{2}$$
Hence equation of tangents are, $$(y-3a)=1(x-2a)$$ and $$(y-3a)=\cfrac{1}{2}(x-2a)$$
$$\Rightarrow \displaystyle x-y+a= 0$$ and $$ x-2y+4a= 0$$
Question 4
1 / -0

The normal drawn at the point $$\displaystyle P\left ( at_{1}^{2},2at_{1} \right )$$ on the parabola meets the curve again at$$\displaystyle Q\left ( at_{2}^{2},2at_{2} \right ).$$ then $$\displaystyle t_{2} =?$$

A
$$\displaystyle t_{2}= -t_{1}-\dfrac{2}{t_{1}}$$

B
$$\displaystyle t_{2}= -t_{1}+\dfrac{2}{t_{1}}$$

C
$$\displaystyle t_{2}= +t_{1}-\dfrac{2}{t_{1}}$$

D
$$\displaystyle t_{2}= +t_{1}+\dfrac{2}{t_{1}}$$

Solution

The normal at $$\displaystyle t_{1}$$ is easily found to be $$\displaystyle y= -t_{1}x+2at_{1}+at_{1}^{3}$$It passes through the point $$\displaystyle \left ( at_{2}^{2},2at_{2} \right )$$
$$\displaystyle \therefore 2at_{2}= -t_{1}at_{2}^{2}+2at_{1}+at_{1}^{3}$$or $$\displaystyle 2a\left ( t_{2}-t_{1} \right )= -at_{1}\left ( t_{2}^{2}-t_{1}^{2} \right )$$
$$\displaystyle \therefore 2= -t_{1}\left ( t_{2}+t_{1} \right ) \because t_{2}\neq t_{1}$$
$$\displaystyle \therefore t_{2}= -t_{1}-\frac{2}{t_{1}}$$

Ans: A
Question 5
1 / -0

The slope of the tangent to the curve $$\displaystyle y=-x^{3}+3x^{2}+9x-27$$ is maximum when x equals.

A
$$1$$

B
$$3$$

C
$$\dfrac 12$$

D
$$-\dfrac 12$$

Solution

Given, $$\displaystyle y=-x^{3}+3x^{2}+9x-27$$
Slope at any point on thegiven curve is $$m= \cfrac{dy}{dx}=-3x^2+6x+9$$
Now for maximum value of slope we must have $$\cfrac{dm}{dx}=0=-6x+6\Rightarrow x = 1$$
Question 6
1 / -0

Find the tangents and normal to the curve $$y(x-2)(x-3)-x+7=0,$$ at point (7,0) are

A
$$x-20y-7=0,$$ $$20x+y-140=0$$.

B
$$x+20y-7=0,$$ $$20x-y-140=0$$.

C
$$ 7x-20y-1=0,$$ $$20x+7y-100=0$$.

D
$$7x+20y-1=0,$$ $$20x-7y-100=0$$.

Solution

Given curve is, $$y(x-2)(x-3)-x+7=0\Rightarrow y(x^2-5x+6)-x+7=0$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}(x^2-5x+6)+y(2x-5)-1=0$$
putting (7,0) to get slope of the tangent
$$\cfrac{dy}{dx}(7^2-5.7+6)+0(2x-5)-1=0\Rightarrow \cfrac{dy}{dx}=\cfrac{1}{20}$$
$$\Rightarrow$$ slope of tangent is $$m = \cfrac{1}{20}$$ and slope of normal is, $$m'=-\cfrac{1}{m}=-20$$
Hence required lines are $$x-20y-7=0,$$ and $$20x+y-140=0$$.
Question 7
1 / -0

Find the distance between the point $$(1,1)$$ and the tangent to the curve $$\displaystyle y=e^{2x}+x^{2}$$ drawn from the point where the curve cuts $$y$$-axis

A
$$\displaystyle \frac{\sqrt3}{\sqrt{5}}$$

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

C
$$\displaystyle \frac{2}{\sqrt{5}}$$

D
$$\displaystyle \frac{\sqrt2}{\sqrt{5}}$$

Solution

Clearly the the point on the $$y$$-axis is $$(0,1)\equiv P$$ (say)
Now $$\cfrac{dy}{dx}=2e^{2x}+2x$$
Thus slope at $$P$$ is $$m=\left(\cfrac{dy}{dx}\right )_{(0,1)}=2$$
Thus required tangent is $$(y-1)=2(x-0)\Rightarrow 2x-y+1=0$$
Hence, distance of $$(1,1)$$ from this line is $$=\left |\cfrac{2\times 1-1+1}{\sqrt{2^2+1^2}}\right |=\cfrac{2}{\sqrt{5}}$$
Question 8
1 / -0

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

A
$$(1, 1)$$

B
$$(0, 1)$$

C
$$(1, 0)$$

D
$$(0,0)$$

Solution

Given, $$\displaystyle y-e^{xy}+x=0$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}-e^{xy}(y+x\cfrac{dy}{dx})+1=0$$
$$\Rightarrow \cfrac{dy}{dx}=\cfrac{1-ye^{xy}}{xe^{xy}-1}$$
Thus for vertical tangent, $$\cfrac{dy}{dx}=\infty$$
$$\Rightarrow xe^{xy}-1=0$$
Hence required point is $$(1,0)$$
Question 9
1 / -0

Find the equation of the tangent to the curve $$\displaystyle y=x^{2}+1$$ at the point $$(1,2).$$

A
$$2y=x$$

B
$$y=2x$$

C
$$y+x=2$$

D
$$y+2x=0$$

Solution

Given $$y=x^2+1$$

$$y'=2x$$ at point $$(1,2)$$

$$\Rightarrow y'=2$$

The equation of tangent $$y-2=2(x-1)$$

$$\Rightarrow y=2x$$

Ans: B
Question 10
1 / -0

The point on the curve $$\displaystyle y=x^{2}-3x+2$$ at which the tangent is perpendicular to the line $$y = x$$ is -

A
$$(0, 2)$$

B
$$(1, 0)$$

C
$$(-1, 6)$$

D
$$(2, -2)$$

Solution

Let the point be $$P(a,b)$$
Now the given curve is $$y=x^2-3x+2$$
Differentiating w.r.t $$x$$
$$\cfrac{dy}{dx}=2x-3$$
Thus slope of tangent at P is $$=\left(\cfrac{dy}{dx}\right)_{(a,b)}=2a-3$$
But given the tangent is perpendicular to line $$y=x$$
$$\Rightarrow 2a-3=-1\Rightarrow a=1$$
Also the point P lies on the given curve,
$$b=a^2-3a+2=0$$
Therefore, the point P is $$(1,0)$$
Hence, option 'B' is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 3

Result

Tangents and its Equations Test 3

Score

-

Rank

-

SHARING IS CARING

Question 1

$$x+y=3$$

$$x-y=3$$

$$2x+y=4$$

$$x+2y=5$$

Solution

Question 2

$$x -2 y + 5 = 0$$

$$x -4 y + 3 = 0$$

$$x - y + 5 = 0$$

$$2x - y + 6 = 0$$

Solution

Question 3

$$\displaystyle x-y+a= 0, x-2y+4a= 0$$

$$\displaystyle x-y-a= 0, x-2y-4a= 0$$

$$\displaystyle x+y+2a= 0, x-2y+a= 0$$

$$\displaystyle x+y-2a= 0, x+2y-4a= 0$$

Solution

Question 4

$$\displaystyle t_{2}= -t_{1}-\dfrac{2}{t_{1}}$$

$$\displaystyle t_{2}= -t_{1}+\dfrac{2}{t_{1}}$$

$$\displaystyle t_{2}= +t_{1}-\dfrac{2}{t_{1}}$$

$$\displaystyle t_{2}= +t_{1}+\dfrac{2}{t_{1}}$$

Solution

Question 5

$$1$$

$$3$$

$$\dfrac 12$$

$$-\dfrac 12$$

Solution

Question 6

$$x-20y-7=0,$$ $$20x+y-140=0$$.

$$x+20y-7=0,$$ $$20x-y-140=0$$.

$$ 7x-20y-1=0,$$ $$20x+7y-100=0$$.

$$7x+20y-1=0,$$ $$20x-7y-100=0$$.

Solution

Question 7

$$\displaystyle \frac{\sqrt3}{\sqrt{5}}$$

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

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

$$\displaystyle \frac{\sqrt2}{\sqrt{5}}$$

Solution

Question 8

$$(1, 1)$$

$$(0, 1)$$

$$(1, 0)$$

$$(0,0)$$

Solution

Question 9

$$2y=x$$

$$y=2x$$

$$y+x=2$$

$$y+2x=0$$

Solution

Question 10

$$(0, 2)$$

$$(1, 0)$$

$$(-1, 6)$$

$$(2, -2)$$

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.