Tangents and its Equations Test 14

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Tangents and its Equations Test 14

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Weekly Quiz Competition

Question 1
1 / -0

The slope of the tangent to the curve $$y= \sqrt {4-x^2}$$ at the point where the ordinate and the abscissa are equal is

A
-1

B
1

C
0

D
none of these

Solution

Given equation of curve is $$y= \sqrt {4-x^2}$$
$$y^{2}=4-x^{2}$$
$$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{x}{y}$$
Let $$P(x_1,y_1)$$ be the point on the curve such that $$x_1=y_1$$
So, slope of tangent to curve at P is $$-1$$.
Question 2
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
none of these

Solution

Given equation of curve is $$x+y=e^{xy}$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{y-1}{1-xe^{xy}}$$
Since, the tangent is parallel to y-axis,
$$\displaystyle\frac{dy}{dx}=\frac{1}{0}$$
$$\Rightarrow xe^{xy}=1$$
Also, $$y=0$$ on the line parallel to y-axis
$$\Rightarrow x=1$$
Hence, the required point is (1,0)
Question 3
1 / -0

If the parabola $$y = ax^2 - 6x + b$$ passes through $$(0, 2)$$ and has its tangent at $$x = \dfrac{3}{2}$$ parallel to the $$x-$$axis then

A
$$a = 2, b = -2$$

B
$$a = 2, b = 2$$

C
$$a = - 2, b = 2$$

D
$$a = - 2, b = - 2$$

Solution

Given equation of parabola is
$$y = ax^2 - 6x + b$$
Since it passes through $$(0, 2)$$
$$2 = 0 + b$$
$$\Rightarrow b = 2$$
Also, $$\displaystyle \frac{dy}{dx} = 2 ax - 6$$
So, slope of tangent at $$\displaystyle (x=\frac {3}{2}) = 2a \left ( \frac{3}{2} \right ) - 6$$
$$= 3a - 6 $$
Since the tangent at $$x=\dfrac{3}{2}$$ is parallel to $$x-$$axis
$$\Rightarrow 3a-6=0$$
$$\Rightarrow a = 2$$
Hence, $$ a=2,b=2$$
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
$$\left (-\displaystyle \frac {22}{9}, -\displaystyle \frac {2}{9}\right )$$

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

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

D
none of these

Solution

Given equation of curve is
$$2x^2 +y^2=12$$
$$\dfrac{dy}{dx}=-\dfrac{2x}{y}$$
Slope of tangent at (2,2) is -2.
Slope of normal at (2,2) is $$\displaystyle\dfrac{1}{2}$$
Equation of normal at (2,2) is
$$y-2=\dfrac{1}{2}(x-2)$$
$$\Rightarrow 2y-x=2$$
Here , we can solve by checking options.
So ,option A satisfies above equation.
Hence $$(-\dfrac{22}{9},-\dfrac{2}{9})$$ is the required point.
Question 5
1 / -0

At what points of curve $$y= \displaystyle \frac {2}{3} x^3 + \displaystyle \frac{1}{2} x^2 $$, the tangent makes equal angle with the axis

A
$$\left( \displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 5 }{ 24 } \right) $$ and $$\left( -1,-\displaystyle \frac { 1 }{ 6 } \right) $$

B
$$\left(\displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 4 }{ 9 } \right)$$ and $$\left( -1,0 \right) $$

C
$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 1 }{ 7 } \right) $$and $$\left( -3,\displaystyle \frac { 1 }{ 2 } \right) $$

D
$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 4 }{ 47 } \right)$$ and $$\left( -1,-\displaystyle \frac { 1 }{ 3 } \right) $$

Solution

Given equation of curve is
$$y= \dfrac {2}{3} x^3 + \dfrac{1}{2} x^2 $$
$$\dfrac{dy}{dx}=2x^{2}+x$$
Since ,the tangent makes equal angle with x-axis,
$$\dfrac{dy}{dx}=1$$
$$\Rightarrow 2x^{2}+x-1=0$$
$$\Rightarrow x=-1,\dfrac{1}{2}$$
At $$ x=-1\Rightarrow y=-\dfrac{1}{6}$$
At $$ x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{24}$$
Question 6
1 / -0

If $$x+ 4y=14$$ is a normal to the curve $$y^2=\alpha x ^3-\beta $$ at $$(2,3)$$, then the value of $$\alpha+\beta$$ is

A
$$9$$

B
$$-5$$

C
$$7$$

D
$$-7$$

Solution

Given equation of curve is
$$y^2=\alpha x ^3-\beta $$
Since, it passes through (2,3)
$$4=8\alpha -\beta$$
$$\displaystyle \frac{dy}{dx}=\frac{3\alpha x^2}{2y}$$
Slope of tangent at (2,3) is $$2\alpha$$
Slope of normal at $$\displaystyle (2,3)= -\frac{1}{2\alpha}$$
Given equation of normal is $$x+4y=14$$
Slope of normal is $$-\cfrac{1}{4}$$
$$\Rightarrow \alpha =2$$
$$\Rightarrow \beta=7$$
Hence $$\alpha+\beta =9$$
Question 7
1 / -0

If a variable tangent to the curve $$x^2y=c^3$$ makes intercepts $$a , b$$ on $$X$$-axes and $$Y$$- axes, respectively, then the value of $$a^2b$$ is

A
$$27c^3$$

B
$$\displaystyle \frac {4}{27}c^3$$

C
$$\displaystyle \frac {27}{4}c^3$$

D
$$\displaystyle \frac {4}{9} c^3$$

Solution

Let a variable point$$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ on the curve $${ x }^{ 2 }y={ c }^{ 3 }$$
Then slope $$\displaystyle\frac { dy }{ dx } =-\frac { 2{ y }_{ 1 } }{ { x }_{ 1 } } $$
So, equation of tangent is
$$\displaystyle y-{ y }_{ 1 }=-\frac { 2{ y }_{ 1 } }{ { x }_{ 1 } } \left( x-{ x }_{ 1 } \right) $$
$$\Rightarrow y{ x }_{ 1 }-2x{ y }_{ 1 }=3{ x }_{ 1 }{ y }_{ 1 }$$
Using $${ { x }_{ 1 } }^{ 2 }{ y }_{ 1 }={ c }^{ 3 }$$
Equation of tangent becomes $$\displaystyle y{ x }_{ 1 }^{ 2 }+\frac { 2{ c }^{ 3 }x }{ { x }_{ 1 } } =3{ c }^{ 3 }$$
Therefore, X intercept is $$\displaystyle a=\frac { 3{ x }_{ 1 } }{ 2 } $$
And Y intercept is $$\displaystyle b=\frac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } } $$
Now $$\displaystyle { a }^{ 2 }b={ \left( \frac { 3{ x }_{ 1 } }{ 2 } \right) }^{ 2 }\left( \frac { 3{ c }^{ 3 } }{ { { x }_{ 1 } }^{ 2 } } \right) =\frac { 27{ c }^{ 3 } }{ 4 } $$
Hence, option 'C' is correct.
Question 8
1 / -0

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

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

B
$$ax+by=1$$

C
$$ax-by=1$$

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

Solution

Given equation of curve is
$$y=be^{-x/a}$$
Let $$P(0,y_1)$$ be the point where tangent to curve crosses y-axis .
$$\Rightarrow y_1=b$$
So, the point P is (0,b)
$$\dfrac{dy}{dx}=-\dfrac{y}{a}$$
Slope of tangent at P(0,b) is $$-\dfrac{b}{a}$$
Equation of tangent through P(0,b) is
$$y-b=-\dfrac{b}{a}(x-0)$$
$$\Rightarrow bx+ay=ab$$
$$\dfrac {x}{a}+\dfrac{y}{b} =1$$
Question 9
1 / -0

The point(s) at each of which the tangents to the curve $$\displaystyle y = x^3 - 3x^2 - 7x + 6$$ cut off on the positive semi axis $$OX$$ a line segment half that on the negative semi axis $$OY$$, then the co-ordinates of the point(s) is/are give by:

A
$$(-1, 9)$$

B
$$(3, -15)$$

C
$$(1, -3)$$

D
none

Solution

$$y={ x }^{ 3 }-3{ x }^{ 2 }-7x+6\\ \dfrac { dy }{ dx } =3{ x }^{ 2 }-6x-7$$
Let coordinate of X be $$\left( a,0 \right) $$ then coordinate of Y is $$\left( 0,-2a \right) $$
Therefore slope of XY is $$\cfrac { 0+2a }{ a-0 } =2$$
Equating slope $$3{ x }^{ 2 }-6x-7=2\\ 3{ x }^{ 2 }-6x-9=0$$
Solving we get
$$x=-1,y=9\quad \\ x=3,y=-15$$
Hence, option 'B' is correct.
Question 10
1 / -0

A line L is perpendicular to the curve $$\displaystyle y = \dfrac {x^2}{4} - 2$$ at its point P and passes through (10, -1). The coordinates of the point P are

A
(2, -1)

B
(6, 7)

C
(0, -2)

D
(4, 2)

Solution

Let point $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ on curve $$ y=\cfrac { { x }^{ 2 } }{ 4 } -2$$
Slope of tangent is $$\dfrac { dy }{ dx } =\dfrac { { x }_{ 1 } }{ 2 } $$
So slope of perpendicular line is $$-\dfrac { 2 }{ { x }_{ 1 } } $$ ...(1)
That perpendicular line also passes through $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$\left( 10,-1 \right) $$
Slope of line from these point is $$\dfrac { { y }_{ 1 }+1 }{ { x }_{ 1 }-10 } $$ ...(2)
Equating (1) and (2) and solving we get,
$${ x }_{ 1 }=4$$ and $${ y }_{ 1 }=2$$
Thus point is $$\left( 4,2 \right) $$
Hence, option 'D' is correct.

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Tangents and its Equations Test 14

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Tangents and its Equations Test 14

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SHARING IS CARING

Question 1

-1

1

0

none of these

Solution

Question 2

(0,1)

(1,0)

(1,1)

none of these

Solution

Question 3

$$a = 2, b = -2$$

$$a = 2, b = 2$$

$$a = - 2, b = 2$$

$$a = - 2, b = - 2$$

Solution

Question 4

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

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

$$(-2,-2)$$

none of these

Solution

Question 5

$$\left( \displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 5 }{ 24 } \right) $$ and $$\left( -1,-\displaystyle \frac { 1 }{ 6 } \right) $$

$$\left(\displaystyle \frac { 1 }{ 2 } ,\displaystyle \frac { 4 }{ 9 } \right)$$ and $$\left( -1,0 \right) $$

$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 1 }{ 7 } \right) $$and $$\left( -3,\displaystyle \frac { 1 }{ 2 } \right) $$

$$\left(\displaystyle \frac { 1 }{ 3 } ,\displaystyle \frac { 4 }{ 47 } \right)$$ and $$\left( -1,-\displaystyle \frac { 1 }{ 3 } \right) $$

Solution

Question 6

$$9$$

$$-5$$

$$7$$

$$-7$$

Solution

Question 7

$$27c^3$$

$$\displaystyle \frac {4}{27}c^3$$

$$\displaystyle \frac {27}{4}c^3$$

$$\displaystyle \frac {4}{9} c^3$$

Solution

Question 8

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

$$ax+by=1$$

$$ax-by=1$$

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

Solution

Question 9

$$(-1, 9)$$

$$(3, -15)$$

$$(1, -3)$$

none

Solution

Question 10

(2, -1)

(6, 7)

(0, -2)

(4, 2)

Solution

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