Tangents and its Equations Test 17

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Tangents and its Equations Test 17

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Weekly Quiz Competition

Question 1
1 / -0

If the tangent to the curve $$\sqrt{x}+\sqrt{y}=\sqrt{a}$$ at any points on it cuts the axes $$OX$$ and $$OY$$ at $$P$$ and $$Q$$ respectively then $$OP+OQ$$ is

A
$$2a$$

B
$$a$$

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

D
none of these

Solution

Let point $$\left( p,q \right) $$ lies on curve $$\sqrt { x } +\sqrt { y } =\sqrt { a } $$
therefore $$\sqrt { p } +\sqrt { q } =\sqrt { a } $$
Now slope $$\cfrac { dy }{ dx } =-\sqrt { \cfrac { p }{ q } } $$
Equation of tangent is $$y-q=-\sqrt { \cfrac { p }{ q } } (x-p)$$
X intercept = OP = $$\sqrt { p } (\sqrt { p } +\sqrt { q } )$$
Y intercept = OQ = $$\sqrt { q } (\sqrt { p } +\sqrt { q } )$$
Hence,
OP+OQ $$=\sqrt { a } \sqrt { a } =a$$
Question 2
1 / -0

The curve $$\displaystyle \frac{x^{n}}{a^{n}}+\frac{y^{n}}{b^{n}}=2$$ touches the line $$\displaystyle \frac{x}{a}+\frac{y}{b}=2$$ at the point

A
$$(b, a)$$

B
$$(a, b)$$

C
$$(1, 1)$$

D
$$\displaystyle \left ( \frac{1}{a}, \frac{1}{b} \right )$$

Solution

Slope of curve$$\cfrac { { x }^{ n } }{ { a }^{ n } } +\cfrac { { y }^{ n } }{ { b }^{ n } } =2$$ is
$$\cfrac { dy }{ dx } =-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } $$
Slope of tangent $$\cfrac { { x } }{ { a } } +\cfrac { { y } }{ { b } } =2$$ is
$$\cfrac { dy }{ dx } =-\cfrac { { b } }{ { a } } $$
Equating both slopes, we get
$$-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } =-\cfrac { { b } }{ { a } } $$
Only $$x=a\quad y=b$$ satisfy the above equation.
Hence, option 'B' is correct.
Question 3
1 / -0

The equation of tangent to the curve $$y=be^{-x/a}$$ where it cuts the $$y$$-axis is

A
$$\displaystyle \frac{x}{a}+\frac{y}{b}=1$$

B
$$\displaystyle \frac{x}{a}+\frac{y}{b}=-1$$

C
$$\displaystyle \frac{x}{a}-\frac{y}{b}=1$$

D
none of these

Solution

Let the point be $$\left( 0,k \right) $$
Substituting this in equation of curve $$y=b{ e }^{ -\cfrac { x }{ a } }$$
we get $$k=b$$ so point is $$\left( 0,b \right) $$
Slope $$\cfrac { dy }{ dx } =-\cfrac { b }{ a } $$
And equation of curve is
$$y-b=-\cfrac { b }{ a } (x-0)\\ \cfrac { x }{ a } +\cfrac { y }{ b } =1$$
Question 4
1 / -0

The normal to the curve $$2x^{2}+y^{2}=12$$ at the point $$(2, 2)$$ cuts the curve again at

A
$$\displaystyle \left ( -\frac{22}{9}, -\frac{2}{9} \right )$$

B
$$\displaystyle \left ( \frac{22}{9}, \frac{2}{9} \right )$$

C
$$(-2, -2)$$

D
none of these

Solution

For curve $$2{ x }^{ 2 }+{ y }^{ 2 }=12$$
Slope of normal is $$-\cfrac { dx }{ dy } =\cfrac { 2y }{ 4x } =\cfrac { 1 }{ 2 } $$
And equation of normal is
$$y-2=\cfrac { 1 }{ 2 } (x-2)\\ 2y-x=2$$
Solving it with equation of curve, we get
$$y=2$$ and $$y=-\cfrac { 2 }{ 9 } $$
Which gives
$$x=2$$ and $$x=-\cfrac { 22 }{ 9 } $$
Hence coordinate of other point is
$$\left( -\cfrac { 22 }{ 9 } ,-\cfrac { 2 }{ 9 } \right) $$
Hence, option 'A' is correct.
Question 5
1 / -0

The area bounded by the axes of reference and the normal to $$y=\log_{e}x$$ at the point $$(1, 0)$$ is

A
1 unit$$^{2}$$

B
2 unit$$^{2}$$

C
$$\displaystyle \frac{1}{2}$$ unit$$^{2}$$

D
none of these

Solution

For curve $$y=\log _{ e }{ x } $$
Slope of normal at point $$(1,0)$$ is
$$-\cfrac { dx }{ dy } =-x=-1$$
Equation of normal is
$$y-0=-1(x-1)\\ x+y=1$$
Therefore X intercept is $$\\ X=1$$
And Y intercept is $$Y=1$$
Hence area $$=\cfrac { 1 }{ 2 } \times 1\times 1=\cfrac { 1 }{ 2 } $$
Question 6
1 / -0

If the line joining the points $$(0, 3)$$ and $$(5, -2)$$ is the tangent to the curve $$\displaystyle y=\frac{c}{x+1}$$ then the value of $$c$$ is

A
$$1$$

B
$$-2$$

C
$$4$$

D
none of these

Solution

Slop of line joining the points $$(0,3)\quad and\quad (5,2)$$ is given by
$$\dfrac { 3+2 }{ 0-5 } =-1$$

This is equal to the slop of tangent on the curve and that is given by

$$\dfrac { dy }{ dx } =\dfrac { -c }{ { \left( x+1 \right) }^{ 2 } } \\ \dfrac { dy }{ dx } =-1\\ \Rightarrow c={ \left( x+1 \right) }^{ 2 }...................(1)$$\

Equation of line joining the points $$(0,3)\quad and\quad (5,2)$$ is given by

$$(y-3)=-1(x-0)$$

Solving equation of tangent and the curve for point of intersection

$$\dfrac { c }{ x+1 } +x=3$$.............(2)

Solving (1) and (2)

$$x=1$$

Putting this in (2), we get c=4
Question 7
1 / -0

The sum of the intercepts made on the axes of coordinates by any tangent to the curve $$\sqrt{x}+\sqrt{y}=2$$ is equal to

A
$$4$$

B
$$2$$

C
$$8$$

D
none of these

Solution

Let point $$\left( p,q \right) $$ lies on the curve $$\sqrt { x } +\sqrt { y } =2$$ , such that $$\sqrt { p } +\sqrt { q } =2$$
Slope is $$\cfrac { dy }{ dx } =-\sqrt { \cfrac { p }{ q } } $$
And equation of tangent is $$y-q=-\sqrt { \cfrac { p }{ q } } (x-p)$$
Therefore X intercept is $$\sqrt { p } \left( \sqrt { p } +\sqrt { q } \right) $$
abd Y intercept is $$\sqrt { q } \left( \sqrt { p } +\sqrt { q } \right) $$
Hence sum of intercept is $$\sqrt { p } \left( \sqrt { p } +\sqrt { q } \right) +\sqrt { q } \left( \sqrt { p } +\sqrt { q } \right) \\ =2\left( \sqrt { p } +\sqrt { q } \right) =4$$
Question 8
1 / -0

The triangle formed by the tangent to the curve $$\displaystyle f(x)=x^{2}+bx-b$$ at the point $$(1, 1)$$ and the coordinates axes, lies in the first quadrant. If its area is $$2$$ then the value of b is

A
-1

B
3

C
-3

D
1

Solution

$$f'(x) = 2x+b$$
$$\Rightarrow m = 2+b$$
So equation of tangent at $$(1,1)$$ is
$$(y-1) = (b+2)(x-1)$$
intercept on the axes are $$(b+1)/(b+2)$$ and $$-(b+1)$$
$$\Rightarrow \mid -\dfrac{1}{2}\dfrac{(b+1)}{(b+2)} (b+1)\mid = 2$$
'$$\Rightarrow \mid \dfrac{(b+1)^2}{b+2}\mid = 4$$
$$\Rightarrow b = -3 $$ is only solution
Question 9
1 / -0

Angle between the tangents to the curve $$y= x^{2}-5x+6$$ at the points $$(2,0)$$ and $$\left ( 3,0 \right )$$ is

A
$$\displaystyle \frac{\pi}{2}$$

B
$$\displaystyle \frac{\pi}{3}$$

C
$$\displaystyle \frac{\pi}{6}$$

D
$$\displaystyle \frac{\pi}{4}$$

Solution

Given equation of curve $$\displaystyle y=x^{2}-5x+6$$

$$\displaystyle \Rightarrow \frac{dy}{dx}=2x-5$$

Slope of tangent to the curve at $$(2,0)$$ is

$$\displaystyle \left(\frac{dy}{dx}\right)_{(2,0)}=2(2)-5=-1=m_{1}$$

Slope of tangent to the curve at $$(3,0)$$ is

$$\displaystyle \left(\frac{dy}{dx}\right)_{(3,0)}=2(3)-5=1=m_{2}$$

Since $$\displaystyle m_{1}m_{2}=-1$$

$$\therefore $$ Angle between the tangents to the curve at $$(2,0)$$ and $$(3, 0)$$ is $$\displaystyle \dfrac{\pi}{2}$$
Question 10
1 / -0

The number of tangents to the curve $$\displaystyle y= e^{\left | x \right |}$$ at the point $$(0,1)$$ is

A
2

B
1

C
4

D
0

Solution

For $$x>0$$,
$$y=e^{x}$$.
$$\dfrac{dy}{dx}=e^{x}$$.
Now
$$\dfrac{dy}{dx}_{x=0}=1$$.
Hence
$$y-1=1(x-0)$$
Or
$$x-y+1=0$$ is the required equation of tangent.
Similarly for $$x<0$$
$$y=e^{-x}$$.
$$\dfrac{dy}{dx}=-e^{-x}$$.
Now
$$\dfrac{dy}{dx}_{x=0}=-1$$.
Hence
$$y-1=-1(x-0)$$
Or
$$x+y-1=0$$ is the required equation of tangent.
Hence at $$x=0$$ we will have 2 tangents to the curve $$y=e^{|x|}$$ which are mutually perpendicular.

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Re-Attempt Weekly Quiz Competition

Tangents and its Equations Test 17

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Tangents and its Equations Test 17

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SHARING IS CARING

Question 1

$$2a$$

$$a$$

$$\displaystyle \frac{1}{2}a$$

none of these

Solution

Question 2

$$(b, a)$$

$$(a, b)$$

$$(1, 1)$$

$$\displaystyle \left ( \frac{1}{a}, \frac{1}{b} \right )$$

Solution

Question 3

$$\displaystyle \frac{x}{a}+\frac{y}{b}=1$$

$$\displaystyle \frac{x}{a}+\frac{y}{b}=-1$$

$$\displaystyle \frac{x}{a}-\frac{y}{b}=1$$

none of these

Solution

Question 4

$$\displaystyle \left ( -\frac{22}{9}, -\frac{2}{9} \right )$$

$$\displaystyle \left ( \frac{22}{9}, \frac{2}{9} \right )$$

$$(-2, -2)$$

none of these

Solution

Question 5

1 unit$$^{2}$$

2 unit$$^{2}$$

$$\displaystyle \frac{1}{2}$$ unit$$^{2}$$

none of these

Solution

Question 6

$$1$$

$$-2$$

$$4$$

none of these

Solution

Question 7

$$4$$

$$2$$

$$8$$

none of these

Solution

Question 8

-1

3

-3

1

Solution

Question 9

$$\displaystyle \frac{\pi}{2}$$

$$\displaystyle \frac{\pi}{3}$$

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{\pi}{4}$$

Solution

Question 10

2

1

4

0

Solution

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