Tangents and its Equations Test 52

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Tangents and its Equations Test 52

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Weekly Quiz Competition

Question 1
1 / -0

The coordinates of the points(s) at which the tangents to the curve $$\displaystyle y=x^{3}-3x^{2}-7x+6$$ cut the positive semi axis OX a line segment half that on the negative semi axis OY is/are given by

A
$$(-1, 9)$$

B
$$(3, -15)$$

C
$$(1, -3)$$

D
none

Solution

Let the point of tangency be
$$(x_{1}+y_{1})$$
Then the equation of tangent is
$$\dfrac{y-y_{1}}{x-x_{1}}=3x_{1}^{2}-6x_{1}-7$$
when
$$y=0,x=x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7}$$
when
$$x=0,y=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$-2(x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$-2(\dfrac{x_{1}(3x_{1}-6x_{1}-7)-y_{1}}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$2=3x_{1}-6x_{1}-7$$
$$3x_{1}-6x_{1}-9=0$$
$$x_{1}=-1 or 3$$
when
$$x_{1}=-1,y_{1}=(-1)^{3}-3(-1)^{2}-7(-1)+6=9$$
and the y-intercept is
$$9-(-1)[3(-1)^{2}-6(-1)-7]=11> 0$$
when
$$x_{1}=3,y_{1}=(3)^{3}-3(3)^{2}-7(3)+6=-15$$
and the y-intercept is
$$-15-(3)[3(3)^{2}-6(3)-7]=-21< 0$$
So,the point of tangency is $$(-1,9)$$, this is the correct answer.
Question 2
1 / -0

The tangent to the curve $$y=e^{x}$$ drawn at the point $$\left ( c,e^{c} \right )$$ intersects the line joining the points $$(c -1,e^{c-1})$$ and $$(c +1,e^{c+1}) $$

A
on the left of $$x = c$$

B
on the right of $$x = c$$

C
at no paint

D
at all points

Solution

Given equation of curve
$$y=e^x$$
$$\dfrac{dy}{dx}=e^x$$
Slope of tangent at point $$(c,e^c)$$ is $$e^c$$
Equation of tangent at $$(c,e^{c})$$ is
$$y-e^{c}=e^{c}(x-c)$$ ....(1)

Equation of straight line joining $$A\left ( c+1,e^{c+1} \right )$$ and $$B\left ( c-1,e^{c-1} \right )$$ is

$$\displaystyle y-e^{c+1}=\dfrac{e^{c+1}-e^{c-1}}{2}\left ( x-c-1 \right )$$ .....(2)

Subtracting (2) from (1), we get
$$\displaystyle e^{c}\left ( e-1 \right )=e^{c}\left [ \left ( x-c \right )-\dfrac{1}{2} \left ( e-e^{-1} \right )\left ( x-c \right )+\dfrac{1}{2}\left ( e-e^{-1} \right )\right ]$$

$$\displaystyle \Rightarrow \dfrac{1}{2}\left ( e+e^{-1} \right )-1=\left ( x-c \right )\left [ 1-\dfrac{1}{2}\left ( e-e^{-1} \right ) \right ]$$

$$\displaystyle \Rightarrow x-c=\dfrac{e+e^{-1}-2}{2-e+e^{-1}}< 0$$

$$[ \because e+e^{-1}>2$$ and $$2+e^{-1}-e<0 ]$$
$$\Rightarrow x<c$$
Thus the two lines meet to the left of $$x = c$$.
Question 3
1 / -0

For the curve $${x}^{2}+4xy+8{y}^{2}=64$$ the tangents are parallel to the $$x$$-axis only at the points

A
$$(0,2\sqrt { 2 } )$$ and $$(0,-2\sqrt { 2 } )$$

B
$$(8,-4)$$ and $$(-8,4)$$

C
$$(8\sqrt { 2 } ,-2\sqrt { 2 } )$$ and $$(-8\sqrt { 2 } ,2\sqrt { 2 } )$$

D
$$(9,0)$$ and $$(-8,0)$$

Solution

Given curve is $${x}^{2}+4xy+8{y}^{2}=64.....(i)$$
On differentiating w.r.t $$x$$ ,we get
$$2x+4(y+x\cfrac{dy}{dx})+16y\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$2x+4y+(4x+16y)\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$\cfrac{dy}{dx}=-\cfrac{(x+2y)}{2(x+4y)}$$
Since, tangent are parallel to $$x$$-axis only
ie., $$\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$-\cfrac{(x+2y)}{2(x+4y)}=0$$
$$\Rightarrow$$ $$x+2y=0.....(ii)$$
Now, on putting the values of $$x$$ from eqs. $$(i)$$ in $$(ii)$$ we get
$$4{y}^{2}-8{y}^{2}+8{y}^{2}=64$$
$$\Rightarrow$$ $${y}^{2}=16$$
$$\Rightarrow$$ $$y=\pm 4$$
from eq. $$(ii)$$
When $$y=4,x=-8$$
and when $$y=-4, x=8$$
Hence, required points are $$(-8,4)$$ and $$(8,-4)$$
Question 4
1 / -0

The equation of the normal to the curve $$y(1+{x}^{2})=2-x$$ where the tangent crosses $$x$$-axis is

A
$$5x-y-10=0$$

B
$$x-5y-10=0$$

C
$$5x+y+10=0$$

D
$$x+5y+10=0$$

Solution

Given curve can be written as, $$y=\dfrac{2-x}{1+x^2}$$

To get the point where the above curve will cross the x-axis, put $$y=0$$

$$\Rightarrow 0=\dfrac{2-x}{1+x^2}\Rightarrow x=2$$

Thus the required point is $$(2,0)$$

Now differentiate the given curve using quotient rule,
$$\dfrac{dy}{dx}=\dfrac{(1+x^2)(-1)-(2-x)(2x)}{(1+x^2)^2}$$

$$\quad =\dfrac{x^2-4x-1}{(1+x^2)^2}$$

Thus the slope of a tangent to this curve at point $$(2,0)$$ is

$$m=\dfrac{dy}{dx}\bigg|_{(2,0)}=\dfrac{2^2-4(2)-1}{(1+2^2)^2}=-\dfrac{5}{25}=-\dfrac{1}{5}$$

Therefore the slope of normal at this point will be $$5$$

Hence required equation of normal is given by, $$(y-0)=5(x-2)$$
$$\Rightarrow 5x-y-10=0$$
Question 5
1 / -0

If $$y = 4x - 5$$ is a tangent to the curve $$y^{2} = px^{3} + q$$ at $$(2, 3)$$, then

A
$$p = 2, q = -7$$

B
$$p = -2, q = 7$$

C
$$p = -2, q = -7$$

D
$$p = 2, q = 7$$

Solution

Since, $$(2, 3)$$ lies on the curve $$y^{2} = px^{3} + q$$
Therefore, $$9 = 8p + q ..... (i)$$
$$y^{2} = px^{3} + q$$
$$\Rightarrow 2y \dfrac {dy}{dx} = 3px^{2}$$
$$\dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$$
$$\Rightarrow \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$$
Since, $$y = 4x - 5$$ is tangent to $$y^{2} = px^{3} + q$$ at $$(2, 3)$$. Therefore,
$$\therefore \left (\dfrac {dy}{dx}\right )_{(2, 3)} =$$ Slope of the line $$y = 4x - 5$$
$$\Rightarrow 2p = 4 \Rightarrow p = 2$$
Putting $$p = 2$$ in Eq. (i), we get $$q = -7$$.
Question 6
1 / -0

The curve given by $$x+y={ e }^{ xy }$$ has a tangent parallel to the y-axis at the point

A
$$(0,1)$$

B
$$(1,0)$$

C
$$(1,1)$$

D
$$(-1,-1)$$

Solution

Given curve is $$x+y={ e }^{ xy }\quad $$
On differentiating w.r.t $$x$$ we get
$$1+\cfrac { dy }{ dx } ={ e }^{ xy }\left\{ y+x\cfrac { dy }{ dx } \right\} $$
$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { y{ e }^{ xy }-1 }{ 1-x{ e }^{ xy } } $$
Since, tangent is parallel to y-axis
Here, $$\cfrac { dy }{ dx } =\infty \Rightarrow 1-x{ e }^{ xy }=0$$
$$\Rightarrow 1-x(x+y)=0$$
This holds for $$x=1,y=0$$
Question 7
1 / -0

The sum of intercepts of the tangent to the curve $$\sqrt { x } +\sqrt { y } =\sqrt { a } $$ upon the coordinates axes is

A
$$2a$$

B
$$a$$

C
$$2\sqrt { 2 } a$$

D
None of these

Solution

Let $$\sqrt { x } +\sqrt { y } =\sqrt { a } $$
diff. curt.
$$\cfrac { 1 }{ 2\sqrt { x } } +\cfrac { 1 }{ 2\sqrt { y } } .\cfrac { dx }{ dy } =0$$
$$\cfrac { dx }{ dy } =-\sqrt { \cfrac { y }{ x } } $$ (slope of tangent)
To find points,
Let $${ x }^{ \circ }=c$$
$${ y }^{ \circ }={ (\sqrt { a } -\sqrt { c } ) }^{ 2 }$$
So, equation of tangent ,
$$\cfrac { y-{ (\sqrt { a } -\sqrt { c } ) }^{ 2 } }{ x-c } =-\sqrt { \cfrac { y }{ x } } \left( =\cfrac { \sqrt { c } -\sqrt { a } }{ \sqrt { c } } \right) \quad \longrightarrow (1)$$
Intercept of x,
put $$y=0$$
$$x=\sqrt { ac } \quad \longrightarrow (2)$$
For intercept of y,
put $$x=0$$ in equation(1)
We get, $$y=a-\sqrt { ac } \quad \longrightarrow (3)$$
Sum of both (2)+(3)
$$a-\sqrt { a } c+\sqrt { a } c$$
$$\Rightarrow a$$
Question 8
1 / -0

Suppose that the equation $$f(x)={x}^{2}+bx+c=0$$ has two distinct real roots $$\alpha,\beta$$. The angle between the tangent to the curve $$y=f(x)$$ at the point $$\left( \cfrac { \alpha +\beta }{ 2 } ,f\left( \cfrac { \alpha +\beta }{ 2 } \right) \right) $$ and the positive direction of the x-axis is

A
$${ 0 }^{ o }$$

B
$${ 30 }^{ o }$$

C
$${ 60 }^{ o }$$

D
$${ 90 }^{ o }$$

Solution

$$f\left( x \right) ={ x }^{ 2 }+bx+c$$
Slope of $$y=f\left( x \right) $$
$$\cfrac { dy }{ dx } =2x+b$$
$${ \left( \cfrac { dy }{ dx } \right) }_{ \cfrac { \alpha +\beta }{ 2 } ,f\left( \cfrac { \alpha +\beta }{ 2 } \right) }=2\times \cfrac { \alpha +\beta }{ 2 } +b$$
from $$f\left( x \right) ={ x }^{ 2 }+bx+c$$ we know $$\alpha +\beta =-b$$
$$\therefore \left( \cfrac { dy }{ dx } \right) =-b+b=0\quad \left[ \tan { 0° } =0=\cfrac { dy }{ dx } \right] $$
Thus, angle between tangent and x-axis$$=0°$$
Question 9
1 / -0

If the slope of the curve $$y=\cfrac { ax }{ b-x } $$ at the point $$(1,1)$$ is $$2$$, then the values of $$a$$ and $$b$$ are respectively

A
$$1,-2$$

B
$$-1,2$$

C
$$1,2$$

D
None of these

Solution

$$y=\cfrac { ax }{ b-x } $$
Slope= $$\cfrac { dy }{ dx } $$
So,$$\cfrac { dy }{ dx } $$= $$\cfrac { a(b-x)-(-1)(ax) }{ { (b-1) }^{ 2 } } \quad \longrightarrow (1)$$
for $$(1,1)$$
put in equation of curve-
$$1=\cfrac { a }{ (b-1) } \quad \Rightarrow (b-1)=a\quad \longrightarrow (2)$$
From slop after putting (1,1)
We get,
$$\\ 2=\cfrac { a(b-1)+a }{ { (b-1) }^{ 2 } } $$
put $$(b-1)=a$$ from equation(2)
$$2=\cfrac { { a }^{ 2 }+a }{ { a }^{ 2 } } \quad \quad \\ \Rightarrow 2{ a }^{ 2 }={ a }^{ 2 }+a\\ \quad \Rightarrow a(a-1)=0\\ \quad \Rightarrow a=0,1$$
put in (2),
We get
$$b=1,2$$
In option only one value is given
$$\therefore a=1,b=2$$
Question 10
1 / -0

The two curves $$x^{3} - 3xy^{2} + 2 = 0$$ and $$3x^{2} y - y^{3} = 2$$

A
Touch each other

B
Cut at right angle

C
Cut at angle $$\dfrac {\pi}{3}$$

D
Cut at angle $$\dfrac {\pi}{4}$$

Solution

Consider the equation $$x^3-3xy^2+2=0$$ and differentiate it with respect to $$x$$:

$$3x^{ 2 }-3y^{ 2 }-6xy\frac { dy }{ dx } =0\\ \Rightarrow 6xy\dfrac { dy }{ dx } =3x^{ 2 }-3y^{ 2 }\\ \Rightarrow \dfrac { dy }{ dx } =\dfrac { 3x^{ 2 }-3y^{ 2 } }{ 6xy } \\ \Rightarrow \dfrac { dy }{ dx } =\frac { x^{ 2 }-y^{ 2 } }{ 2xy } \quad .....(1)$$

Now, consider the equation $$3x^2y-y^3-2=0$$ and differentiate it with respect to $$x$$:

$$6xy+3x^{ 2 }\dfrac { dy }{ dx } -3y^{ 2 }\dfrac { dy }{ dx } =0\\ \Rightarrow 6xy+\dfrac { dy }{ dx } (3x^{ 2 }-3y^{ 2 })=0\\ \Rightarrow \dfrac { dy }{ dx } =\dfrac { -6xy }{ 3x^{ 2 }-3y^{ 2 } } \\ \Rightarrow \dfrac { dy }{ dx } =\dfrac { -2xy }{ x^{ 2 }-y^{ 2 } } \quad .....(2)$$

Multiply 1 and 2 as follows:

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

In general, if the product of two slopes is equal to $$-1$$ then the lines are perpendicular, therefore, the two given curves are perpendicular to each other.

Hence, the two curves $$x^3-3xy^2+2=0$$ and $$3x^2y-y^3-2=0$$ cut each other at right angle.

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Tangents and its Equations Test 52

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Tangents and its Equations Test 52

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SHARING IS CARING

Question 1

$$(-1, 9)$$

$$(3, -15)$$

$$(1, -3)$$

none

Solution

Question 2

on the left of $$x = c$$

on the right of $$x = c$$

at no paint

at all points

Solution

Question 3

$$(0,2\sqrt { 2 } )$$ and $$(0,-2\sqrt { 2 } )$$

$$(8,-4)$$ and $$(-8,4)$$

$$(8\sqrt { 2 } ,-2\sqrt { 2 } )$$ and $$(-8\sqrt { 2 } ,2\sqrt { 2 } )$$

$$(9,0)$$ and $$(-8,0)$$

Solution

Question 4

$$5x-y-10=0$$

$$x-5y-10=0$$

$$5x+y+10=0$$

$$x+5y+10=0$$

Solution

Question 5

$$p = 2, q = -7$$

$$p = -2, q = 7$$

$$p = -2, q = -7$$

$$p = 2, q = 7$$

Solution

Question 6

$$(0,1)$$

$$(1,0)$$

$$(1,1)$$

$$(-1,-1)$$

Solution

Question 7

$$2a$$

$$a$$

$$2\sqrt { 2 } a$$

None of these

Solution

Question 8

$${ 0 }^{ o }$$

$${ 30 }^{ o }$$

$${ 60 }^{ o }$$

$${ 90 }^{ o }$$

Solution

Question 9

$$1,-2$$

$$-1,2$$

$$1,2$$

None of these

Solution

Question 10

Touch each other

Cut at right angle

Cut at angle $$\dfrac {\pi}{3}$$

Cut at angle $$\dfrac {\pi}{4}$$

Solution

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