Tangents and its Equations Test 32

Result

Tangents and its Equations Test 32

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

Equation of normal drawn to the graph of the function defined as $$f(x)=\dfrac{\sin\,x^2}{x},x \neq0 $$ and $$f(0)=0$$ at the origin is

A
$$x+y=0$$

B
$$x-y=0$$

C
$$y=0$$

D
$$x=0$$

Solution

Let the equation of line is $$y=mx$$
$$f(x)=\dfrac{\sin x^2}{x}$$
Let $$g(x)=\sin x^2$$
For $$\left | x \right |< \delta $$:
$$g(x)=\displaystyle \sin x^2=x^2-\dfrac{(x^2)^3}{3!}+\dfrac{(x^2)^5}{5!}-...=\sum_{0}^{\infty }(-1)^n\dfrac{(x^2)^{(2n+1)}}{(2n+1)!} $$
$$f(x)=\dfrac{x^2-\dfrac{(x^2)^3}{3!}+\dfrac{(x^2)^5}{5!}-...}{x}$$
$$f(x)=x-\dfrac{x^5}{3!}+\dfrac{x^9}{5!}...$$
$$\dfrac{df(x)}{dx}=1-\dfrac{5x^4}{3!}+\dfrac{9x^8}{5!}...$$
$$m_1=1-\dfrac{5x^4}{3!}+\dfrac{9x^8}{5!}...$$ $$(m_1:$$ slope of tangent line at $$x=0)$$
$$m_1=1$$
$$m=\dfrac{-1}{m_1}$$

$$\therefore m=-1$$

Equation of normal is:
$$y+x=0$$

Hence, Option A is correct.
Question 2
1 / -0

The slope of the tangent to the curve $$y=\int _{ 0 }^{ x }{ \frac { dt }{ 1+{ t }^{ 3 } } } $$ at the point where $$x=1$$ is

A
$$\frac { 1 }{ 4 } $$

B
$$\frac { 1 }{ 3 } $$

C
$$\frac { 1 }{ 2 } $$

D
$$1$$

Solution

$$\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 3 } } $$
$$m=\left( \frac { dy }{ dx } \right) x=1=\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 } $$
Question 3
1 / -0

A tangent PT is drawn to the circle $$x^2+y^2=4$$ at the point $$P(\sqrt{3}, 1)$$. A straight line L, perpendicular to PT is a tangent to the circle $$(x-3)^2+y^2=1$$. $$(1)$$ A possible equation of L is?

A
$$x-\sqrt{3}y=1$$

B
$$x+\sqrt{3}y=1$$

C
$$x-\sqrt{3}y=-1$$

D
$$x+\sqrt{3}y=5$$

Solution

Slope of tangent to $$x^{2}+y^{2}=4$$ at $$P(\sqrt{3},1)$$ is given by differentiating given equation w.r.t. x
$$\dfrac{dy}{dx}=\dfrac{-x}{y}=-\sqrt{3}$$
So slope of line L $$=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}$$
Let equation of L be $$x-\sqrt3y+c=0$$
As L is tangent to $$(x-3)^{2}+y^{2}=1$$
So perpendicular distance of L from its center should be equal to its radius i.e. $$1$$.
Centre $$(3,0)$$
$$\dfrac{|3+c|}{\sqrt4}=1$$
$$\Rightarrow c=-1 or -5$$
So equation of L is $$x-\sqrt3y=1$$
or
$$x-\sqrt3y=5$$
Hence option A is correct.
Question 4
1 / -0

The points of the curve $$y={ x }^{ 3 }+x-2$$ at which its tangent are parallel to the straight line $$y=4x-1$$ are

A
$$\left( 2,7 \right) ,\left( -2,-11 \right) $$

B
$$\left( 0,-2 \right) ,\left( { 2 }^{ 1/3 },{ 2 }^{ 1/3 } \right) $$

C
$$\left( { -2 }^{ 1/3 },{ -2 }^{ 1/3 } \right) \left( 0,-4 \right) $$

D
$$\left( 1,0 \right) ,\left( -1,-4 \right) $$

Solution

Given
$$y={ x }^{ 3 }+x-2...(i)$$

$$y=4x-1....(ii)$$

Slope of tangent to the curve (i)

$$\cfrac { dy }{ dx } =3{ x }^{ 2 }+1$$

slope of tangent at point $$\left( \alpha ,\beta \right) $$ is

$$\quad { \left| \cfrac { dy }{ dx } \right| }_{ \left( \alpha ,\beta \right) }^{ }=3{ \alpha }^{ 2 }+1...(iii)$$

Given, tangent of curve (i) is parallel to line (ii)

$$\therefore$$ Slope of line (ii) is $$4$$

$$\therefore$$ From Eq(iii) we get

$$3{ \alpha }^{ 2 }+1=4\Rightarrow \alpha =\pm 1$$

$$\therefore \left( \alpha ,\beta \right) $$ lie on curve (i)

$$\beta ={ \left( \pm 1 \right) }^{ 2 }+\left( \pm 1 \right) -2\Rightarrow \beta =0,-4\quad $$

$$\therefore$$ Points are $$(1,0)$$ and $$(-1,-4)$$
Question 5
1 / -0

The equation of the other normal to the parabola $${y}^{2}=4ax$$ which passes through the intersection of those at $$(4a,-4a)$$ and $$(9a,-6a)$$ is:

A
$$5x-y+115a=0$$

B
$$5x+y-135a=0$$

C
$$5x-y-115a=0$$

D
$$5x+y+115=0$$

Solution

There is a slight misprint and the equation of the parabola should be $${ y }^{ 2 }=4ax$$.
Any point on the parabola is of the form P ($$at^2$$,$$2at$$).
Now, differentiating the equation of the parabola,
$$\quad { 2y\dfrac { dy }{ dx } }=4a\\ \Rightarrow \frac { dy }{ dx } =\frac { 2a }{ y }$$
So, Slope of the Normal$$=\dfrac { -y }{ 2a } =\dfrac { -2at }{ 2a } =-t$$

Hence, Equation of Normal at P ($$at^2$$,$$2at$$) is:
$$y-2at=-t\left( x-a{ t }^{ 2 } \right)$$
$$\Rightarrow a{ t }^{ 3 }+2at-xt-y=0$$ ...(1)
In this cubic equation, sum of roots =0
Hence, we can say $${ t }_{ 1 }+{ t }_{ 2 }+{ t }_{ 3 }=0$$
Now, at point 1 ($$4a$$,$$-4a$$),
$$-4a=2a{ t }_{ 1 }$$
$$\Rightarrow { t }_{ 1 }=-2$$
Similarly, at point 2 ($$9a$$,$$-6a$$),
$$2a{ t }_{ 2 }=-6a$$
$$\Rightarrow { t }_{ 2 }=-3$$

Now as $${ t }_{ 1 }+{ t }_{ 2 }+{ t }_{ 3 }=0$$,
we get, $${ t }_{ 3 }=5$$
Putting $${ t }_{ 3 }=5$$,
From the equation of normal in (1) above, we get
$$a\left( 125 \right) +5\left( -x+2a \right) -y=0$$
$$\Rightarrow 135a=5x+y$$
or, $$5x+y-135a=0$$ ....(B)
Question 6
1 / -0

Let $$f\left( x \right) =\begin{cases} -{ x }^{ 2 }\ \ \ \ \ , x<0 \\ { x }^{ 2 }+8,x\ge 0 \end{cases}$$ Equation of tangent line touching both branches of $$y=f\left( x \right)$$ is

A
$$y=4x+1$$

B
$$y=4x+4$$

C
$$y=x+4$$

D
$$y=x+1$$
Question 7
1 / -0

The area of the triangle formed by the positive x-axis, the tangent and normal to the curve $${ x }^{ 2 }+{ y }^{ 2 }=16{ a }^{ 2 }$$ at the point $$\left( 2\sqrt { 2 } a,2\sqrt { 2 } a \right) $$ is

A
$${a}^{2}$$

B
$$16{a}^{2}$$

C
$$4{a}^{2}$$

D
$$8{a}^{2}$$

Solution

Point on the curve is $$(2\sqrt{2}a,2\sqrt{2}a)$$

Equation of tangent at this point is $$x_1x+y_1y=16a^2$$

here $$x_1 = 2\sqrt{2}a$$ and $$y_1=2\sqrt{2}a$$

$$Area = \dfrac{1}{2} \times{base}\times{height}$$

Point of intersection of tangent to the x axis is $$\dfrac{16a^2}{2\sqrt{2}a}$$
So,
Base = $$4\sqrt{2}a$$

Height =$$2\sqrt{2}a $$

$$Area = \dfrac{1}{2}\times{4\sqrt{2}a}\times{2\sqrt{2}a}$$

$$Area = 8a^2$$
Question 8
1 / -0

A point P moves such that sum of the slopes of the normal drawn from it to the hyperbola $$xy=16$$ is equal to the sum of the ordinates of the feet of the normal. Let 'P' lies on the curve C, then.
The equation of 'C' is?

A
$$x^2=4y$$

B
$$x^2=16y$$

C
$$x^2=12y$$

D
$$y^2=8x$$

Solution

Let P(h,k) be the point in the plane of hyperbola $$xy=16=4^2$$

The equation of the normal at a point $$(4t, \dfrac {4} {t})$$ to the hyperbola $$xy=4^2$$ is

$$xt^3-yt-4t^4+4=0$$

if if passes through $$P(h,k)$$ then,

$$ht^3-yt-4t^4+4=0$$

$$ \Rightarrow 4t^4-ht^3+kt-4=0$$

this is a fourth degree equation in t. So, it gives 4 values of t, $$t_1,t_2,t_3,t_4$$ corresponding to each value of t there is a point on hyperbola such that the normal passes through $$P(h,k)$$.

let the 4 points be $$ A(ct_1, \dfrac{c}{t_1}) $$,$$ B(ct_2, \dfrac{c}{t_2}) $$,$$ C(ct_3, \dfrac{c}{t_3}) $$ and $$ D(ct_4, \dfrac{c}{t_4}) $$

such that normal at these points pass through $$P(h,k)$$.

Since $$t_1,t_2,t_3,t_4$$ are roots of equation (1).

Therefore,
$$t_1+t_2+t_3+t_4 = \dfrac{h}{c}=\dfrac{h}{4}$$.......(2)

$$ \sum t_1t_2 =0 $$.......(3)

$$ \sum t_1t_2t_3 = - \dfrac{k}{4} $$.........(4)

$$ t_1t_2t_3t_4 =-1 $$.......(5)

It s given that the sum of the slopes of the normals at A,B,C and D is equal to the sum of the coordinates of these points.

Therefore
$$t_1^2+t_2^2+t_3^2+t_4^2 = \dfrac{c}{t_1}+\dfrac{c}{t_3}+\dfrac{c}{t_3}+\dfrac{c}{t_4}=\dfrac{4}{t_1}+\dfrac{4}{t_2}+\dfrac{4}{t_3}+\dfrac{4}{t_4}$$

$$(t_1+t_2+t_3+t_4 )^2-2\sum t_1t_2=4\left ( \dfrac {\sum t_1t_2t_3}{t_1t_2t_3t_4} \right )$$

$$\Rightarrow \dfrac{h^2}{4}-2\times 0 = 4 \times \left ( \dfrac{\dfrac {-k}{4}} {-1} \right ) $$

$$\Rightarrow \dfrac{h^2}{4} = k $$

$$\Rightarrow h^2 = 16k$$

so the locus of $$(h,k)$$ is $$x^2=16y$$, which is a parabola
Question 9
1 / -0

The equation of normal $${ x }^{ 2/3 }+{ y }^{ 2/3 }={ a }^{ 2/3 }$$ at $$\left( \cfrac { a }{ 2\sqrt { 2 } } ,\cfrac { a }{ 2\sqrt { 2 } } \right) $$ is ______ .

A
$$2x+y=0$$

B
$$y=1$$

C
$$x=0$$

D
$$x=y$$

Solution

To get the slope of $${ x }^{ \cfrac { 2 }{ 3 } }+{ y }^{ \cfrac { 2 }{ 3 } }={ a }^{ \cfrac { 2 }{ 3 } }$$
We differentiate w.r.to $$x$$
$$\quad \therefore \cfrac { 2 }{ 3 } { x }^{ -\cfrac { 1 }{ 3 } }+\cfrac { 2 }{ 3 } { y }^{ -\cfrac { 1 }{ 3 } }\cfrac { dy }{ dx } =0$$
$$\therefore \cfrac { dy }{ dx } =-\left( \cfrac { { x }^{ -\cfrac { 1 }{ 3 } } }{ { y }^{ \cfrac { 2 }{ 3 } } } \right) $$
$$\therefore \cfrac { dy }{ dx } =-{ \left( \cfrac { y }{ x } \right) }^{ \cfrac { 1 }{ 3 } }$$
$$\therefore$$ Slope of tangent at point
$$P\left( \cfrac { a }{ 2\sqrt { 2 } } ,\cfrac { a }{ 2\sqrt { 2 } } \right) $$
$$\therefore$$ Slope $$\quad m={ \left( \cfrac { dy }{ dx } \right) }_{ \cfrac { a }{ 2\sqrt { 2 } } ,\cfrac { a }{ 2\sqrt { 2 } } }\quad $$
$$\quad \therefore m={ \left( \cfrac { dy }{ dx } \right) }_{ \cfrac { a }{ 2\sqrt { 2 } } ,\cfrac { a }{ 2\sqrt { 2 } } }-{ \left( \cfrac { \cfrac { a }{ 2\sqrt { 2 } } }{ \cfrac { a }{ 2\sqrt { 2 } } } \right) }^{ \cfrac { 1 }{ 3 } }\quad $$
$$=-{ (1) }^{ \cfrac { 1 }{ 3 } }=-1$$
Hence slope of tangent $$m=-1$$
$$\therefore$$ Slope of normal $$=1$$
$$\therefore$$ we get equation of normal using equation $$y-{ y }_{ 1 }=m(x-{ x }_{ 1 })\quad $$
$$\therefore y-\cfrac { a }{ 2\sqrt { 2 } } =(1)\left( x-\cfrac { a }{ 2\sqrt { 2 } } \right) $$
$$\therefore y-\cfrac { a }{ 2\sqrt { 2 } } =x-\cfrac { a }{ 2\sqrt { 2 } } $$
$$\therefore$$ $$y=x$$
Question 10
1 / -0

The tangent to $$\left( a{ t }^{ 2 },2at \right) $$ is perpendicular to X-axis at _____ point $$t\in R$$.

A
$$(4a,4a)$$

B
$$(a,2a)$$

C
$$(0,0)$$

D
$$(a,-2a)$$

Solution

Here $$\left( x,y \right) =\left( a{ t }^{ 2 },2at \right) $$
$$\therefore x=a{ t }^{ 2 };y=2at$$
$$\therefore \cfrac { dx }{ dt } =2at;\cfrac { dy }{ dt } =2a$$
$$\therefore \cfrac { dy }{ dx } \cfrac { dy/dt }{ dx/dt } =\cfrac { 2a }{ 2at } =\cfrac { 1 }{ t } $$
Since, tangent is perpendicular to X-axis slope is undefined
$$\therefore$$ $$t=0$$
$$\therefore t=0$$
$$\therefore \left( a{ t }^{ 2 },2at \right) =\left( 0,0 \right) $$
Thus, tangent is perpendicular to X-axis at $$(0,0)$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 32

Result

Tangents and its Equations Test 32

Score

-

Rank

-

SHARING IS CARING

Question 1

$$x+y=0$$

$$x-y=0$$

$$y=0$$

$$x=0$$

Solution

Question 2

$$\frac { 1 }{ 4 } $$

$$\frac { 1 }{ 3 } $$

$$\frac { 1 }{ 2 } $$

$$1$$

Solution

Question 3

$$x-\sqrt{3}y=1$$

$$x+\sqrt{3}y=1$$

$$x-\sqrt{3}y=-1$$

$$x+\sqrt{3}y=5$$

Solution

Question 4

$$\left( 2,7 \right) ,\left( -2,-11 \right) $$

$$\left( 0,-2 \right) ,\left( { 2 }^{ 1/3 },{ 2 }^{ 1/3 } \right) $$

$$\left( { -2 }^{ 1/3 },{ -2 }^{ 1/3 } \right) \left( 0,-4 \right) $$

$$\left( 1,0 \right) ,\left( -1,-4 \right) $$

Solution

Question 5

$$5x-y+115a=0$$

$$5x+y-135a=0$$

$$5x-y-115a=0$$

$$5x+y+115=0$$

Solution

Question 6

$$y=4x+1$$

$$y=4x+4$$

$$y=x+4$$

$$y=x+1$$

Question 7

$${a}^{2}$$

$$16{a}^{2}$$

$$4{a}^{2}$$

$$8{a}^{2}$$

Solution

Question 8

$$x^2=4y$$

$$x^2=16y$$

$$x^2=12y$$

$$y^2=8x$$

Solution

Question 9

$$2x+y=0$$

$$y=1$$

$$x=0$$

$$x=y$$

Solution

Question 10

$$(4a,4a)$$

$$(a,2a)$$

$$(0,0)$$

$$(a,-2a)$$

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.