Tangents and its Equations Test 40

Result

Tangents and its Equations Test 40

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

If the slope of one of the lines represented $${a^3}{x^2} + 2hxy + {b^3}{y^2} = 0$$ be the square of the other, then $$ab(a+b)$$ is equal to:

A
$$2h$$

B
$$-2h$$

C
$$8h$$

D
$$-8h$$

Solution
Question 2
1 / -0

Slope of tangent to the circle $$( x - r ) ^ { 2 } + y ^ { 2 } = r ^ { 2 }$$ at the point $$( x , y )$$ lying on the circle is

A
$$\frac { x } { y - r }$$

B
$$\frac { r - x } { y }$$

C
$$\frac { y ^ { 2 } - x ^ { 2 } } { 2 x y }$$

D
$$\frac { y ^ { 2 } + x ^ { 2 } } { 2 x y }$$

Solution

$$(x-r)^2+y^2=r^2$$

$$\dfrac{d}{dx}(x-r)^2+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(r^2)$$

$$2(x-r)+2y\dfrac{dy}{dx}=0$$

$$x-r+y\dfrac{d}{dx}=0$$

Therefore the slope of the tangent is,

$$\dfrac{dy}{dx}=\dfrac{r-x}{y}$$
Question 3
1 / -0

The slope of the straight line which is both tangent and normal to the curve $$4x^3=27y^2$$ is

A
$$\underline { + } 1$$

B
$$\underline { + } \dfrac { 1 }{ 2 } $$

C
$$\underline { + } \dfrac { 1 }{ \sqrt { 2 } } $$

D
$$\underline { + } \sqrt { 2 } $$

Solution

$$ Given: $$
$$ 4{{x}^{3}}=27{{y}^{2}} $$
$$ differentiating\,w.r.t.x $$
$$ 12{{x}^{2}}=54yy' $$
$$ t $$
$$ The\text{ line is tangent at}\,\text{P}\left( 3{{t}^{2}},2{{t}^{3}} \right)and\,normal\,at\left( 3t_{1}^{2},2t_{1}^{3} \right) $$
$$ \frac{dy}{dx}=\frac{2\left( 9{{t}^{4}} \right)}{9\left( 2{{t}^{3}} \right)}=1 $$
$$ equation\,to\,\tan gent\,P\left( t \right)is $$
$$ y-2{{t}^{3}}=t\left( x-3{{t}^{2}} \right) $$
$$ \Rightarrow tx-y={{t}^{3}}\,\,\,\,\,........\left( i \right) $$
$$ Equation\,to\,normal\,atQ\left( t_{1}^{{}} \right)is $$
$$ y-2t_{1}^{3}=-\frac{1}{t_{1}^{{}}}\left( x-3t_{1}^{2} \right) $$
$$ \Rightarrow x+t_{1}^{{}}y=2t_{1}^{4}+3t_{1}^{2}\,\,\,\,\,\,\,\,...........\left( ii \right) $$
$$ \left( i \right)\,and\left( ii \right)are\,identical $$
$$ \frac{t}{1}=\frac{-1}{t_{1}^{{}}}=\frac{{{t}^{3}}}{2t_{1}^{4}+3t_{1}^{2}} $$
$$ \Rightarrow {{t}^{2}}=\frac{2}{{{t}^{4}}}+\frac{3}{{{t}^{2}}}\,\,\,\,\,\,\,\,or\,\,t_{1}^{{}}=-\frac{1}{t} $$
$$ \Rightarrow {{t}^{6}}=2+3{{t}^{2}} $$
$$ \Rightarrow {{t}^{6}}-3{{t}^{2}}-2=0 $$
$$ \Rightarrow {{t}^{2}}=2 $$
$$ t=\pm \sqrt{2} $$
Question 4
1 / -0

The angle between the curves $$y = \sin x$$ and $$y = \cos x$$ is

A
$$\tan^{-1}(2\sqrt{2})$$

B
$$\tan^{-1}(3\sqrt{2})$$

C
$$\tan^{-1}(3\sqrt{3})$$

D
$$\tan^{-1}(5\sqrt{2})$$

Solution

C₁ : $$y = \sin x$$, C₂ : $$y = \cos x$$.
Equating the $$y$$' s,
$$\sin x = \cos x$$ ∴ $$x = \dfrac{\pi}{4} $$
∴ curves intersect each other at the point P : $$x = \dfrac{\pi}{4}$$.
Now, differentiating w.r.t. $$x$$,
C₁ gives : $$\dfrac{dy}{dx} = \cos x$$
C₂ gives : $$\dfrac{dy}{dx} = - \sin x$$.
Hence slopes $m₁ and m₂ of C₁ and C₂ at P : $$x = \dfrac{π}{4}$$ are
m₁ $$= \cos \dfrac{π}{4} = \dfrac{1}{√2}$$
m₂ =$$ - \sin \dfrac{π}{4} = -\dfrac{1}{√2}$$.
If $$θ$$ is the acute angle between them at $$P$$, then
$$tan θ =\dfrac{ | ( m₁ - m₂ )|}{ |( 1 + m₁m₂ ) |}$$
$$=\dfrac{ | ((\dfrac{1}{√2}) - (-\dfrac{1}{√2}))| }{ |( 1 + (\dfrac{1}{√2})(-\dfrac{1}{√2})) | }$$
$$=\dfrac{ | 2( \dfrac{1}{√2 })|}{ |( \dfrac{(2-1)} { (2)} | }$$
$$= 2√2 $$
∴ $$θ = \tanֿ¹ ( 2√2 ) $$
Question 5
1 / -0

The tangent at any point of the curve $$x={ at }^{ 3 },y={ at }^{ 4 }$$ divides the abscissa of the point of contact in the ratio

A
$$1:4$$

B
$$3:2$$

C
$$1:3$$

D
$$3:1$$

Solution

Given:
$$x = at^2$$
$$y = at^4$$

Then,
$$\Rightarrow t^3 = \dfrac{x}{a}, t^4= \dfrac{y}{a}$$

$$\Rightarrow t = \left(\dfrac{x}{a}\right)^{1/3}\, t = \left(\dfrac{y}{a}\right)^{1/4}$$

$$\Rightarrow \left(\dfrac{x}{a}\right)^{1/3} = \left(\dfrac{y}{a}\right)^{1/4}$$

$$\Rightarrow \left(\dfrac{x}{a}\right)^{\frac{1}{3}\times \frac{4}{12}} = \left(\dfrac{y}{a} \right)^{\frac{1}{4}\times 12}$$

$$\Rightarrow \dfrac{x^4}{a^4} = \dfrac{y^3}{a^2}$$

$$\Rightarrow y^3 = \dfrac{x^4}{a}$$ at $$P(h,k)$$

Also relation $$k^3 = \dfrac{h^4}{a}$$

Now by differentiation of after equation we get

$$3y^2 \dfrac{dy}{dx} = \dfrac{4x^2}{a}$$

$$\dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$$

Now $$M_T = \dfrac{dy}{dx} = \dfrac{4x^3}{3ay^2}$$

Now we will find slope of tangent

$$M_T / P(h,x) = \dfrac{4h^3}{3ax^2}$$

equation of tangent $$P(h,k)$$

$$y - k = \dfrac{4h^3}{3ak^2}(x-h)$$

$$\Rightarrow $$ Let $$y = 0$$

Then, $$-k = \dfrac{4h^3}{3ak^2}(x-h)$$

$$\Rightarrow \dfrac{-3ak^3}{4h^3} = x - h$$

Earlier we found that $$k^3 = \dfrac{h^4}{a}$$

Then $$ax^3 = h^4$$

$$\dfrac{-3h^4}{4h^2} = x-h$$

$$\Rightarrow \dfrac{-3h}{4}+h = x$$

$$\Rightarrow x = \dfrac{h}{4}$$

$$\Rightarrow \dfrac{x}{h} = \dfrac{1}{4}$$

Hence option (a) $$1 : 4$$ is the correct choice
Question 6
1 / -0

The tangent to the curve $$y=e^{2x}$$ at the point $$(0, 1)$$ meets x-axis at?

A
$$(0, 1)$$

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

C
$$(2, 0)$$

D
$$(0, 2)$$

Solution

solution - equation of curve is y=$$e^{2x}$$
tangent to the curve is
$$\frac{dy}{dx}=2e^{2x}$$
$$(\frac{de^{x}}{dx}=e^{x})$$
At points (0,1), the tangent is
$$(\frac{dy}{dx})_{(0,1)}=2e^{2(0)}$$
$$(\frac{dy}{dx})_{(0,1)}=2e^{0}$$ $$[e^{0}=1]$$
$$\frac{dy}{dx}_{0,1}=2$$
when the slope is 2, and solve the tangent
meets y axis the coordinate is 0,
Hence the tangent meets the x axis at (2,0)
Question 7
1 / -0

The tangent to the curve, $$y = xe^{x^2}$$ passing through the point $$(1, e)$$ also passes through the point:

A
$$\left(\dfrac{4}{3}, 2e\right)$$

B
$$(2, 3e)$$

C
$$\left(\dfrac{5}{3}, 2e\right)$$

D
$$(3, 6e)$$

Solution

$$y = xe^{x^2}$$

$$\dfrac{dy}{dx}\left|_{(1,e)} = \left(x\cdot e^{x^2} \cdot 2x+e^{x^2}\right)\right|_{1,e} = 2\cdot e + e = 3e$$

$$T : y - e = 3e (x - 1)$$

$$y = 3ex - 3e + e$$

$$y = (3e) x - 2e$$

Out of the options only option:A

$$\left(\dfrac{4}{3}, 2e\right)$$ lies on it as $$2e=3e\times \dfrac 43-2e\Rightarrow 2e=2e$$
Question 8
1 / -0

Three normals are drawn from the point $$\left(c,0\right)$$ to the curve $${y}^{2}=x.$$If two of the normals are perpendicular to each other,then $$c=$$

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{2}$$

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

D
$$1$$

Solution

$$ Let\,the\text{ equation of normal to }{{\text{y}}^{2}}=4ax\,is $$
$$ y=mx-2am-a{{m}^{2}} $$
$$ \therefore equation\,of\,normal\,for\,{{y}^{2}}=x\,is $$
$$ y=mx-\frac{m}{2}-\frac{1}{4}{{m}^{3}}which\,passes\,through\left( c,0 \right) $$
$$ \therefore 0=m\left( c-\frac{1}{2}-\frac{{{m}^{2}}}{4} \right) $$
$$ \Rightarrow m=0\,\,and\,\frac{{{m}^{2}}}{4}=c-\frac{1}{2} $$
$$ \Rightarrow m=\pm 2\sqrt{c-\frac{1}{2}} $$
$$ \text{which gives a normal as x-axis and for other two normal} $$
$$ c-\frac{1}{2}>0 $$
$$ \Rightarrow c>\frac{1}{2} $$
$$ N\text{ow, if normals are perpendicular}\text{.} $$
$$ \left( 2\sqrt{c-\frac{1}{2}} \right)\left( -2\sqrt{c-\frac{1}{2}} \right)=-1 $$
$$ \Rightarrow c-\frac{1}{2}=\frac{1}{4} $$
$$ \Rightarrow c=\frac{3}{4} $$
Question 9
1 / -0

If the line $$x+y=0$$ touches the curve $$2y^2=\alpha x^2+\beta $$ at $$(1,-1),$$ then $$(\alpha ,\beta )=$$

A
$$(-2,4)$$

B
$$(-1,3)$$

C
$$(4,-2)$$

D
$$(2,0)$$

Solution

Given equation of curve is,
$$2{ y }^{ 2 }=\alpha { x }^{ 2 }+\beta $$

Point $$\left( 1,-1 \right) $$ lies on the curve. Thus, it must satisfy given equation of curve.

$$\therefore 2{ \left( -1 \right) }^{ 2 }=\alpha { \left( 1 \right) }^{ 2 }+\beta $$

$$\therefore 2\left( 1 \right) =\alpha { \left( 1 \right) }+\beta $$

$$\therefore 2=\alpha +\beta $$ (1)

Now, equation of tangent to curve is,
$$x+y=0$$
$$\therefore y=-x$$

Thus, slope of tangent is, $${ m }_{ 1 }=-1$$ (2)

Equation of the curve is,
$$2{ y }^{ 2 }=\alpha { x }^{ 2 }+\beta $$
Differentiate w.r.t. x, we get,

$$2\times 2y\frac { dy }{ dx } =2\alpha x+0$$

$$\therefore 4y\frac { dy }{ dx } =2\alpha x$$

$$\therefore \frac { dy }{ dx } =\frac { 2\alpha x }{ 4y } $$

Slope of curve at $$\left( 1,-1 \right) $$ is,
$${ m }_{ 2 }=\frac { 2\alpha \times 1 }{ 4\times -1 } $$

$${ m }_{ 2 }=\frac { 2\alpha }{ -4 } $$

$$\therefore { m }_{ 2 }=\frac { \alpha }{ -2 } $$ (3)

At point $$\left( 1,-1 \right) $$, slope of tangent and normal will be same.
Thus, from equation (2) and (3),

$$\frac { \alpha }{ -2 } =-1$$

$$\therefore \alpha =2$$

Put this value in equation (1), we get,

$$2+\beta =2$$

$$\therefore \beta =0$$
$$\therefore \left( \alpha ,\beta \right) =\left( 2,0 \right) $$
Question 10
1 / -0

Equation of the tangent at (1, -1) to the curve
$${ x }^{ 3 }-x{ y }^{ 2 }-4{ x }^{ 2 }-xy+5x+3y+1=0$$ is

A
$$x-4y-5=0$$

B
$$x+1=0$$

C
$$y-1=0$$

D
$$y+1=0$$

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 40

Result

Tangents and its Equations Test 40

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2h$$

$$-2h$$

$$8h$$

$$-8h$$

Solution

Question 2

$$\frac { x } { y - r }$$

$$\frac { r - x } { y }$$

$$\frac { y ^ { 2 } - x ^ { 2 } } { 2 x y }$$

$$\frac { y ^ { 2 } + x ^ { 2 } } { 2 x y }$$

Solution

Question 3

$$\underline { + } 1$$

$$\underline { + } \dfrac { 1 }{ 2 } $$

$$\underline { + } \dfrac { 1 }{ \sqrt { 2 } } $$

$$\underline { + } \sqrt { 2 } $$

Solution

Question 4

$$\tan^{-1}(2\sqrt{2})$$

$$\tan^{-1}(3\sqrt{2})$$

$$\tan^{-1}(3\sqrt{3})$$

$$\tan^{-1}(5\sqrt{2})$$

Solution

Question 5

$$1:4$$

$$3:2$$

$$1:3$$

$$3:1$$

Solution

Question 6

$$(0, 1)$$

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

$$(2, 0)$$

$$(0, 2)$$

Solution

Question 7

$$\left(\dfrac{4}{3}, 2e\right)$$

$$(2, 3e)$$

$$\left(\dfrac{5}{3}, 2e\right)$$

$$(3, 6e)$$

Solution

Question 8

$$\dfrac{1}{4}$$

$$\dfrac{1}{2}$$

$$\dfrac{3}{4}$$

$$1$$

Solution

Question 9

$$(-2,4)$$

$$(-1,3)$$

$$(4,-2)$$

$$(2,0)$$

Solution

Question 10

$$x-4y-5=0$$

$$x+1=0$$

$$y-1=0$$

$$y+1=0$$

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.