Tangents and its Equations Test 6

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Tangents and its Equations Test 6

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Weekly Quiz Competition

Question 1
1 / -0

The gradient of the tangent line at the point $$(a cos \alpha, a sin \alpha)$$ to the circle $$x^2 + y^2 = a^2$$, is

A
$$tan (\pi - \alpha)$$

B
$$ tan \alpha$$

C
$$ cot \alpha$$

D
- $$ cot \alpha$$

Solution

Equation of the circle, $$x^2+y^2=a^2$$

At a point $$(a \cos \alpha, a \ sin \alpha)$$

Gradient of radius $$= \dfrac{y_2-y_1}{x_2-x_1}$$

$$= \dfrac{a \ sin \alpha-0}{a \ cos \alpha-0}$$

$$=\ tan \alpha $$

gradient of radius $$\times$$ gradient of tangent = -1

$$\ tan \alpha \times$$ gradient of tangent = -1

$$\Rightarrow $$gradient of tangent$$ = -\ tan \alpha $$

$$\Rightarrow $$gradient of tangent$$ = \ tan (\pi - \alpha)$$
Question 2
1 / -0

If the product of the slope of tangent to curve at $$(x,y)$$ and its y-co-ordinate is equal to the x-co-ordinate of the point, then it represent.

A
circle

B
parabola

C
ellipse

D
rectangular hyperbola

Solution

Let slope of a tangent at point $$(x,y)$$ to the curve is $$=\cfrac { dy }{ dx } $$
$$\therefore y\cfrac { dy }{ dx } =x$$ (given)
$$\therefore ydy=xdx\quad $$
Integrating both side $$\int { y } dy=\int { x } dx$$
$$\therefore \cfrac { { y }^{ 2 } }{ 2 } =\cfrac { { x }^{ 2 } }{ 2 } +c'$$
$$\therefore { y }^{ 2 }-{ x }^{ 2 }+2c'$$
$$\therefore { y }^{ 2 }-{ x }^{ 2 }=c$$
where $$2c'=c$$
Which is a rectangular hyperbola
Question 3
1 / -0

The length of the segment of the tangent line to the curve $$x=a\cos ^{ 3 }{ t } ,y=\sin ^{ 3 }{ t } $$, at any point on the curve cut off by the coordinate axes is

A
$$4a$$

B
$$a$$

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

D
$$2a$$
Question 4
1 / -0

Which one of the following be the gradient of the hyperbola $$xy=1$$ at the point $$\left(t,\dfrac{1}{t}\right)$$

A
$$-\dfrac{1}{t}$$

B
$$-\dfrac{1}{t^2}$$

C
$$\dfrac{1}{t}$$

D
$$-\dfrac{2}{t^2}$$

Solution

Given the hyperbola
$$xy=1$$.
Now differentiate both sides with respect to $$x$$ we get,
$$y+x\dfrac{dy}{dx}=0$$
or, $$\dfrac{dy}{dx}=-\dfrac{y}{x}$$
Now the gradient of the given hyperbola at $$\left(t,\dfrac{1}{t}\right)$$ is
$$\left.\dfrac{dy}{dx}\right|_{\left(t,\dfrac{1}{t}\right)}=-\dfrac{1}{t^2}$$
Question 5
1 / -0

Equation of tangent at that point of the curve $$y = 1 - {e^{\frac{x}{2}}}$$, where it meets y-axis

A
$$x + 2y = 0$$

B
$$2x + y = 0$$

C
$$x - y = 2$$

D
None of these

Solution

Given ,

$$ y=1-{{e}^{\frac{x}{2}}} $$ ……(1)

Differentiate w.r. t x both side we get

$$ \dfrac{dy}{dx}=-{{e}^{\frac{x}{2}}}\times \dfrac{1}{2} $$ …..(2)

Given that graph of equation $$(1)$$ meets y-axis

$$\Rightarrow x=0$$
Put in equation $$(1),$$ we get

$$ y=1-{{e}^{\frac{0}{2}}} $$

$$ y=1-{{e}^{0}} $$

$$ y=1-1 $$

$$ y=0 $$

Put $$y=0$$ in equation $$(1)$$ we get
$$x=0$$

Now, put the value of $$x,y$$ in equation $$(2),$$ we get

$$ {{\left( \dfrac{dy}{dx} \right)}_{\left( 0,0 \right)}}=-{{e}^{\dfrac{0}{2}}}\times \dfrac{1}{2}=-1\times \dfrac{1}{2}=-\dfrac{1}{2} $$

Now required equation is
$$ y-{{y}_{1}}=\dfrac{dy}{dx}\left( x-{{x}_{1}} \right) $$

$$ y-0=-\dfrac{1}{2}\left( x-0 \right) $$

$$ y=-\dfrac{1}{2}x $$

$$ 2y+x=0 $$
This is the required equation of tangent.
Question 6
1 / -0

The slope of the tangent to the curve $$xy+ax-by=0$$ at the point $$(1,1)$$ is $$2$$, then value of $$a$$ and $$b$$ are respectively:

A
$$1,2$$

B
$$2,1$$

C
$$3,5$$

D
None of these

Solution

$$xy+ax-by=0$$
$$(1,1)$$ lies on curve
$$1+a-b=0$$
$$a-b=-1$$ ....... $$(i)$$
$$y+x\dfrac { dy }{ dx } +a-b\dfrac { dy }{ dx } =0$$
$$1+2+a-2b=0$$
$$a-2b=-3$$ ........ $$(ii)$$
Solving $$(i)$$ and $$(ii)$$, we get
$$a=1,b=2$$
Question 7
1 / -0

The Equation of the tangent to the curves $${y^2} = 8x$$ and $$xy = -1$$ is

A
$$3y=9x+2$$

B
$$y=2x+1$$

C
$$2y=x+1$$

D
$$y=x+2$$

Solution

Any tangent to $${y^2} = 8x$$ is
$$y = mx + \dfrac{2}{m}$${equation of tangent of parabola}
it should also be tangent to $$xy=-1$$
so $$x\left( {mx + \dfrac{2}{m}} \right) = - 1$$
$$m^2x^2+2x+m=0$$
$$\therefore D=0 $$
$$4-4m^3=0$$
$$\therefore m=1$$
so tangent $$y = mx + \dfrac{2}{m}$$
$$ \Rightarrow y = x + 2$$
Question 8
1 / -0

If the tangent to the curve $$y=x\log { x } $$ at $$\left( c,f\left( x \right) \right) $$ is parallel to the line-segment joining $$A\left(1,0\right)$$ and $$B\left(e,e\right)$$, then c=...... .

A
$$\dfrac {e-1}{e}$$

B
$$\log { \dfrac { e-1 }{ e } } $$

C
$${ e }^{ \dfrac { 1 }{ 1-e } }$$

D
$${ e }^{ \dfrac { 1 }{ e-1 } }$$

Solution

slope of AB $$\displaystyle =\frac{e}{e-1}$$
$$\displaystyle \frac{dy}{dx}=1+\log x$$
slope at $$(c,f\left(x\right ))$$
$$\displaystyle 1+\log c=\frac{e}{e-1}\implies=e^{\frac{1}{e-1}}$$
Question 9
1 / -0

The values of $$x$$ for which the tangents to the curves $$y=x\cos{x},y=\cfrac{\sin{x}}{x}$$ are parallel to the axis of $$x$$ are roots of (respectively)

A
$$\sin{x}=x,\tan{x}=x$$

B
$$\cot{x}=x,\sec{x}=x$$

C
$$\cot{x}=x,\tan{x}=x$$

D
$$\tan{x}=x,\cot{x}=x$$

Solution

$$y=x\cos x$$

$$\dfrac{dy}{dx}=\cos x-x\sin x=0$$

$$\implies x=\cot x$$

$$y=\dfrac{\sin x}{x}$$

$$\dfrac{dy}{dx}=\dfrac{x\cos x-\sin x}{x^{2}}=0$$

$$\implies x=\tan x$$
Question 10
1 / -0

A curve with equation of the form $$y=a{x}^{4}+b{x}^{3}+cx+d$$ has zero gradient at the point $$(0,1)$$ and also touches the x-axis at the point $$(-1,0)$$ then

A
$$a=3$$

B
$$b=4$$

C
$$c+d=1$$

D
for $$x< -1$$ the curve has a negative gradient

Solution

given $$y=ax^4+bx^3+cx+d$$
$$y_{}{'}=4ax^3+3bx^2+c=0$$ at (0,1)
so $$c=0$$
curve passes through point (-1,0)
so $$0=a-b-c+d=>a+d=b$$
here $$y_{}{'}<0$$ for $$x<-1$$

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 6

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Tangents and its Equations Test 6

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SHARING IS CARING

Question 1

$$tan (\pi - \alpha)$$

$$ tan \alpha$$

$$ cot \alpha$$

- $$ cot \alpha$$

Solution

Question 2

circle

parabola

ellipse

rectangular hyperbola

Solution

Question 3

$$4a$$

$$a$$

$${a}^{2}$$

$$2a$$

Question 4

$$-\dfrac{1}{t}$$

$$-\dfrac{1}{t^2}$$

$$\dfrac{1}{t}$$

$$-\dfrac{2}{t^2}$$

Solution

Question 5

$$x + 2y = 0$$

$$2x + y = 0$$

$$x - y = 2$$

None of these

Solution

Question 6

$$1,2$$

$$2,1$$

$$3,5$$

None of these

Solution

Question 7

$$3y=9x+2$$

$$y=2x+1$$

$$2y=x+1$$

$$y=x+2$$

Solution

Question 8

$$\dfrac {e-1}{e}$$

$$\log { \dfrac { e-1 }{ e } } $$

$${ e }^{ \dfrac { 1 }{ 1-e } }$$

$${ e }^{ \dfrac { 1 }{ e-1 } }$$

Solution

Question 9

$$\sin{x}=x,\tan{x}=x$$

$$\cot{x}=x,\sec{x}=x$$

$$\cot{x}=x,\tan{x}=x$$

$$\tan{x}=x,\cot{x}=x$$

Solution

Question 10

$$a=3$$

$$b=4$$

$$c+d=1$$

for $$x< -1$$ the curve has a negative gradient

Solution

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