Tangents and its Equations Test 9

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Tangents and its Equations Test 9

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Weekly Quiz Competition

Question 1
1 / -0

Equation of the tangent to the curve $$y(x-2)(x-3)-x+7=0$$ at the point where
the curve cuts x-axis is

A
$$x-20y=7$$

B
$$x-20y+7=0$$

C
$$x+20y-7=0$$

D
$$x+20y+7=0$$

Solution

$$y=0$$ (at x-axis)
$$0-x+7=0$$
$$x=7$$
$$y=\dfrac{x-7}{(x-2)(x-3)}$$
$${y}'=\dfrac{(x-2)(x-3)-(x-7)(2x-5)}{(x-2)^{2}(x-3)^{2}}$$
put $$x=7$$
$$\dfrac{5\times 4-(0)}{(5\times 4)^{2}}$$
$${y}'=\dfrac{1}{20}$$
$$(y-0)=\dfrac{1}{20}(x-7)$$
$$x-20y-7=0$$
Question 2
1 / -0

P(1, 1) is a point on the parabola $$y=x^{2}$$ whose vertex is A. The point on the curve at which the tangent drawn is parallel to the chord $$\overline{AP}$$ is

A
$$(\displaystyle \frac{1}{2}, \frac{1}{4})$$

B
$$(\displaystyle \frac{-1}{2}, \frac{1}{4})$$

C
$$(2, 4)$$

D
$$(4, 2)$$

Solution

$$tan \theta =\dfrac{(1-0)}{(1-0)}=1$$
$${y}'=2x$$
$$2x=1$$
$$x=\dfrac{1}{2}, y=\dfrac{1}{4}$$
Question 3
1 / -0

Equation of the tangent to the parabola $$y^{2}=4x+5$$ which is parallel to the line $$y= 2x + 7$$ is

A
$$y = 2x + 3$$

B
$$y = 2x- 3$$

C
$$y = 2x + 5$$

D
$$y = 2x - 5$$

Solution

Differentiating the equation given,we obtain
$$2{y}'y=2$$
$${y}'=2$$
$$\dfrac{2}{y}=2$$
$$y=1$$
$$x=-1$$
$$(y-1)=2(x+1)$$
$$y=2x+3$$
Question 4
1 / -0

For the parabola $$y^{2}=8x$$, tangent and normal are drawn at $$P(2, 4)$$ which meet the axis of the parabola in $$A$$ and $$B$$, then the length of the diameter of the circle through $$A, P, B$$ is

A
$$2$$

B
$$4$$

C
$$8$$

D
$$6$$

Solution

Considering the equation of the parabola
$$y^{2}=8x$$
The axis of symmetry is the positive $$x-$$axis.
By, differentiating with respect to $$x$$, we get
$$2y.y'=8$$
$$y'=\dfrac{4}{y}$$
Hence, $$y'_{(2,4)}=1$$
Thus slope of the tangent at the point of contact $$P(2,4)$$ is $$1$$ while that of normal is $$-1$$.
Hence, equation of the tangent will be
$$\dfrac{y-4}{x-2}=1$$
$$-x+y=2$$ ...(i)
Hence, the tangent meets the x axis at $$(-2,0)$$
Therefore, $$A=(-2,0)$$.

Equation of normal will be
$$\dfrac{y-4}{x-2}=-1$$
$$x+y=6$$ ...(ii)
Hence the normal meets the x axis at $$(6,0)$$.
Thus, $$B=(6,0)$$

Therefore the three points through which the circle passes are
$$(-2,0),(6,0),(2,4)$$
Now let the equation of the circle be
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$
Therefore
$$(-2-h)^{2}+k^{2}=r^{2}$$
$$\rightarrow (2+h)^{2}+k^{2}=r^{2}$$
$$(6-h)^{2}+k^{2}=r^{2}$$
Subtracting {ii} from {i}, we get
$$2h-4=0$$
$$h=2$$
Therefore the equation of the circle reduces to
$$(x-2)^{2}+(y-k)^{2}=r^{2}$$
Now
$$(2-2)^{2}+(4-k)^{2}=r^{2}$$
$$\rightarrow (4-k)^{2}=r^{2}$$
$$(6-2)^{2}+k^{2}=r^{2}$$
$$\rightarrow 16+k^{2}=r^{2}$$
Subtracting i from ii, we get
$$16+k^{2}-(k-4)^{2}=0$$
$$16+(2k-4)(4)=0$$
$$4+2k-4=0$$
$$k=0$$
Thus the centre of the circle lies at $$C=(2,0)$$.
Hence the radius of the circle is
$$CA=CP=CB$$
Now
$$CA=\sqrt{(2-(-2))^{2}+0^{2}}$$
$$=2-(-2)$$
$$=4$$
$$=r$$

Hence, the diameter of the circle is $$8$$ units.
Question 5
1 / -0

If the curves $$y=x^{2}-1,\ y=8x-x^{2}-9$$ touch each other at (2, 3) then equation of the common tangent is

A
$$4x-y=5$$

B
$$4x+y=5$$

C
$$x-4y=5$$

D
$$x+4y=14$$

Solution

$$y_{1}^{1}=2x$$
$$y_{2}^{1}=8-2x$$
at the tangent point slope will be
some.
$$2x=8-2x$$
$$4x=8$$
$$x=2$$
$$y=3$$
$$y^{1}=4$$
$$(y-3)=4(x-2)$$
$$y=4x-5$$
Question 6
1 / -0

The point on the hyperbola $$y = \dfrac {x - 1}{x + 1}$$ at which the tangents are parallel to $$y = 2x + 1$$ are

A
$$(0, -1)$$ only

B
$$(-2, 3)$$ only

C
$$(0, -1)$$, $$(-2, 3)$$

D
$$(-2, 3)$$, $$(5, 4)$$

Solution

$${y}'=\dfrac{(x+1)-(x-1)}{(x+1)^{2}}$$
$${y}'=2 (\therefore$$ parallel to $$y =2x+1)$$
$$\dfrac{2}{(x+1)^{2}}=2$$
$$(x+1)^{2}=1$$
$$x=0$$ or $$x=-2$$
Using the equation of hyperbola, when $$x=0$$, $$y=-1$$.
when $$x=-2$$, $$y=3$$
Question 7
1 / -0

Assertion(A): The tangent to the curve $$y=x^{3}-x^{2}-x+2$$ at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.

A
Both A and R are true R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$y^{1}=3x^{2}-2x-1$$
$$y^{1}|_{x=1}=0=m$$
$$m=0$$
$$\therefore$$ the curve is parallel to x-axis.
Question 8
1 / -0

Assertion A: The curves $$x^{2}=y,\ x^{2}=-y$$ touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.

A
Both A and R are true R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$y'=2x$$
$$y'|_{x=0}=0=m_{1}$$
$$y'=-2x$$
$$y'|_{x=0}=0=m_{2}$$
$$m_{1}=m_{2}$$
$$\therefore$$ both the curve touches each other
Question 9
1 / -0

Observe the following statements for the curve $$y=2.e^{\frac{-x}{3}}$$
I : The slope of the tangent to the curve where it meets y-axis is $$\displaystyle \frac{-2}{3}$$
II:The equation of normal to the curve where it meets y-axis is $$3x+2y+4=0$$.
Which of the above statement is correct

A
only I

B
only II

C
both I and II

D
neither I nor II

Solution

$$y'=\cfrac{-2}{3}e^{-x}$$
$$y'|_{x=0}=\frac{-2}{3}$$
$$y=2,\ x=0$$
$$(y-2)=\dfrac{3}{2}(x-0)$$
$$2y-3x-4=0$$
only 2 is true
Question 10
1 / -0

Match the points on the curve $$2y^{2}=x+1$$ with the slope of normals at those points and choose
the correct answer.
Point
Slope of normal
I : $$(7, 2)$$

$$a){-4\sqrt{2}}$$

II: $$(0, \displaystyle \frac{1}{\sqrt{2}})$$

$$b) -8$$
III : $$(1, 1)$$
$$c) -4$$
IV: $$(3, \sqrt{2})$$

$$d){-2\sqrt{2}}$$

A
$$i-b,ii- d,iii- c,iv- a$$

B
$$i-b,ii- a,iii- d,iv- c$$

C
$$i-b,ii- c,iii- d,iv- a$$

D
$$i-b,ii- d,iii- a,iv- c$$

Solution

$$\displaystyle y'=\frac{1}{4y}=m$$

Slope of normal $$m'=-4y$$

For given points slope is
(1) $$m'_{1}=-4\times 2 ==-8$$
(2) $$m'_{2}=-4\times \dfrac{1}{\sqrt 2}=-2\sqrt{2}$$
(3) $$m'_{3}=-4\times 1=-4$$
(4) $$m'_{4}=-4\times {\sqrt 2}=-4\sqrt{2}$$

Option A is correct answer

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Re-Attempt Weekly Quiz Competition

Point	Slope of normal
I : $$(7, 2)$$	$$a){-4\sqrt{2}}$$
II: $$(0, \displaystyle \frac{1}{\sqrt{2}})$$	$$b) -8$$
III : $$(1, 1)$$	$$c) -4$$
IV: $$(3, \sqrt{2})$$	$$d){-2\sqrt{2}}$$

Tangents and its Equations Test 9

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Tangents and its Equations Test 9

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SHARING IS CARING

Question 1

$$x-20y=7$$

$$x-20y+7=0$$

$$x+20y-7=0$$

$$x+20y+7=0$$

Solution

Question 2

$$(\displaystyle \frac{1}{2}, \frac{1}{4})$$

$$(\displaystyle \frac{-1}{2}, \frac{1}{4})$$

$$(2, 4)$$

$$(4, 2)$$

Solution

Question 3

$$y = 2x + 3$$

$$y = 2x- 3$$

$$y = 2x + 5$$

$$y = 2x - 5$$

Solution

Question 4

$$2$$

$$4$$

$$8$$

$$6$$

Solution

Question 5

$$4x-y=5$$

$$4x+y=5$$

$$x-4y=5$$

$$x+4y=14$$

Solution

Question 6

$$(0, -1)$$ only

$$(-2, 3)$$ only

$$(0, -1)$$, $$(-2, 3)$$

$$(-2, 3)$$, $$(5, 4)$$

Solution

Question 7

Both A and R are true R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 8

Both A and R are true R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 9

only I

only II

both I and II

neither I nor II

Solution

Question 10

$$i-b,ii- d,iii- c,iv- a$$

$$i-b,ii- a,iii- d,iv- c$$

$$i-b,ii- c,iii- d,iv- a$$

$$i-b,ii- d,iii- a,iv- c$$

Solution

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