** NCERT Solutions for Maths Class 12 Exercise 6.3**

Hello Students! Welcome to **maths-formula.com**. In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 6.3**.

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Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

**NCERT Solutions for Maths Class 12 Exercise 6.3** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

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**NCERT Solutions for Maths Class 12 Exercise 6.3** are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

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**NCERT Solutions for Maths Class 12 Exercise 6.1**

**NCERT Solutions for Maths Class 12 Exercise 6.2**

**NCERT Solutions for Maths Class 12 Exercise 6.4**

**NCERT Solutions for Maths Class 12 Exercise 6.5**

**NCERT Solutions for Maths Class 12 Exercise 6.3**

**Maths Class
12 Ex 6.3 Question 1.**

Find the slope of
the tangent to the curve y = 3x^{4} – 4x at x = 4.

**Solution:**

The given curve
is y = 3x^{4} – 4x

∴ dy/dx = 12x^{3} – 4

∴
Slope of the
tangent to the curve = (dy/dx)_{x = 4}

= 12 × 4^{3} – 4 = 764

**Maths Class
12 Ex 6.3 Question 2.**

Find the slope of
the tangent to the curve at
x = 10.

**Maths Class
12 Ex 6.3 Question 3.**

Find the slope of
the tangent to curve y = x^{3} – x + 1 at the point whose
x-coordinate is 2.

**Solution:**

The given curve
is y = x^{3} – x + 1

dy/dx = 3x² – 1

∴
Slope of the
tangent to the curve = (dy/dx)_{x = 2}

= 3 × 2² – 1

= 11

**Maths Class
12 Ex 6.3 Question 4.**

Find the slope of
the tangent to the curve y = x^{3} – 3x + 2 at the point whose
x-coordinate is 3.

**Solution:**

The given curve is y = x^{3} – 3x + 2

dy/dx = 3x² – 3

∴
Slope of the
tangent to the curve = (dy/dx)_{x = 3}

= 3 × 3² –
3

= 24

**Maths Class
12 Ex 6.3 Question 5.**

Find the slope of
the normal to the curve x = a cos^{3} Î¸, y = a sin^{3} Î¸
at Î¸ = Ï€/4.

**Solution:**

The given curve
is x = a cos^{3} Î¸, y = a sin^{3} Î¸

Find the slope of
the normal to the curve x = 1 – a sin Î¸, y = b cos² Î¸ at Î¸ = Ï€/4.

**Solution:**

The given curve
is x = 1 – a sin Î¸, y = b cos² Î¸

Find points at
which the tangent to the curve y = x^{3} – 3x^{2} –
9x + 7 is parallel to the x-axis.

**Solution:**

Differentiating
w.r.t. x; dy/dx = 3(x – 3) (x +
1)

Tangent is parallel to x-axis if the slope of tangent = 0

or dy/dx = 0

⇒
3(x + 3)(x + 1) =
0

⇒
x = –1, 3

When x = –1, y = 12 and when x = 3, y = –20

Hence, the tangent to the given curve are parallel to x-axis at the points (–1,
12), (3, –20)

**Maths Class
12 Ex 6.3 Question 8.**

Find a point on
the curve y = (x – 2)² at which the tangent is parallel to the chord joining
the points (2, 0) and (4, 4).

**Solution:**

The given equation of the curve is y = (x – 2)²

Differentiating w.r.t x, we get dy/dx = 2(x – 2)

The points A and B are (2, 0) and (4, 4), respectively.

Slope of AB = (y_{2} – y_{1})/(x_{2} – x_{1}) =
(4 – 0)/(4 – 2) = 4/2 = 2 … (i)

Slope of the tangent = 2(x – 2) … (ii)

From eq. (i) and (ii), 2(x – 2) = 2

∴ x – 2 = 1 or x = 3

when x = 3, y = (3 – 2)² = 1

∴ The tangent is parallel to the chord AB at (3, 1).

**Maths Class
12 Ex 6.3 Question 9.**

Find the point on
the curve y = x^{3} – 11x + 5 at which the tangent is y = x – 11.

**Solution:**

The given curve
is y = x^{3} – 11x + 5

⇒
dy/dx = 3x² – 11

The slope of tangent line y = x – 11 is 1.

∴ 3x² – 11 = 1

⇒ 3x² = 12

⇒ x² = 4, x = ±2

When x = 2, y = –9 and when x = –2, y = 19

But (–2, 19) does not lie on the curve.

∴ y = x – 11 is the tangent at (2, –9).

**Maths Class
12 Ex 6.3 Question 10.**

Find the equation
of all lines having slope –1 that are tangents to the curve y = 1/(x – 1), x ≠ 1.

**Solution:**

The given curve
is y = 1/(x – 1), x ≠ 1

The point where the tangents to the given curve have the slope -1 are (2,
1) and (0, -1).

Equation of the tangent at (2, 1) is

Y – 1 = -1(x – 2) or x + y = 3

Equation of the tangent at (0, -1) is

Y + 1 = -1(x – 0) or x + y + 1 = 0

Thus, the required lines are x + y = 3 and x + y + 1 = 0.

**Maths Class
12 Ex 6.3 Question 11.**

Find the equation
of all lines having slope 2 which are tangents to the curve y = 1/ (x – 3), x ≠
3.

**Solution:**

The given curve
is y = 1/(x – 3), x ≠ 3

dy/dx = –1/(x
– 3)^{2}

Since slope of the tangent = 2

–1/(x – 3)^{2 }= 2 or (x – 3)^{2
}= –1/2

Which is not
possible as (x – 3)² > 0

Thus, no tangent to y = 1/(x – 3) has slope 2.

**Maths Class
12 Ex 6.3 Question 12.**

Find the
equations of all lines having slope 0 which are tangent to the curve y = 1/(x^{2}
– 2x + 3).

**Solution:**

Let the tangent
at the point (x_{1}, y_{1}) to the curve

**Maths Class
12 Ex 6.3 Question 13.**

Find points on
the curve x^{2}/9 + y^{2}/16 = 1 at which the tangents are

(a) parallel to x-axis

(b) parallel to y-axis

**Solution:**

The equation of
the curve is

Therefore, the
tangents are parallel to y-axis at (3, 0) and (–3, 0).

**Maths Class
12 Ex 6.3 Question 14.**

Find the
equations of the tangent and normal to the given curves at the indicated
points:

(i) y = x^{4} – 6x^{3} + 13x^{2} – 10x +
5 at (0, 5)

(ii) y = x^{4} – 6x^{3} + 13x^{2} – 10x
+ 5 at (1, 3)

(iii) y = x^{3} at (1, 1)

(iv) y = x^{2} at (0, 0)

(v) x = cos t, y = sin t at t = Ï€/4

**Solution:**

The given curve
is y = x^{4} – 6x^{3} + 13x^{2} – 10x +
5

dy/dx = 4x^{3}
– 18x^{2} + 26x – 10

Putting x = 0, dy/dx at (0, 5)
= –10

**Maths Class
12 Ex 6.3 Question 15.**

Find the equation
of the tangent line to the curve y = x^{2} – 2x + 7 which is

(a) parallel to the line 2x – y + 9 = 0

(b) perpendicular to the line 5y – 15x = 13.

**Solution:**

The given curve
is y = x² – 2x + 7 … (i)

dy/dx = 2x – 2 = 2(x – 1)

(a) Slope of the line 2x – y + 9 = 0 is 2

⇒ Slope of tangent = dy/dx = 2(x – 1) = 2

**Maths Class
12 Ex 6.3 Question 16.**

Show that the
tangents to the curve y = 7x^{3} + 11 at the points where x = 2
and x = – 2 are parallel.

**Solution:**

The given curve
is y = 7x^{3} + 11

dy/dx = 21x²

Now m_{1} = slope at x = 2 is (dy/dx)_{x = 2} =
21 × 2² = 84

And m_{2} = slope at x = -2 is (dy/dx)_{x = -2} =
21 × (-2)² = 84

Here, m_{1} = m_{2}

Thus, the tangents
to the given curve at the points where x = 2 and x = –2 are parallel.

**Maths Class
12 Ex 6.3 Question 17.**

Find the points
on the curve y = x^{3} at which the slope of the tangent is equal
to the y-coordinate of the point.

**Solution:**

Let P(x_{1},
y_{1}) be the required point.

The given curve is y = x^{3 } … (i)

**Maths Class
12 Ex 6.3 Question 18.**

For the curve y =
4x^{3} – 2x^{5}, find all the points at which the tangent
passes through the origin.

**Solution:**

Let (x_{1},
y_{1}) be the required point on the given curve y = 4x^{3} –
2x^{5},

then y_{1} =
4x_{1}^{3} – 2x_{1}^{5} … (i)

**Maths Class
12 Ex 6.3 Question 19.**

Find the points
on the curve x^{2} + y^{2} – 2x – 3 = 0 at which the
tangents are parallel to the x-axis.

**Solution:**

The given curve
is x^{2} + y^{2} – 2x – 3 = 0

dy/dx = (1 – x)/y

Tangent is parallel to x-axis, if dy/dx = 0, i.e.
if 1 – x = 0 ⇒ x = 1

Putting x = 1 in (i), we get y = ±2

Hence, the required points are (1, 2), (1, –2).

**Maths Class
12 Ex 6.3 Question 20.**

Find the equation
of the normal at the point (am^{2}, am^{3}) for the curve ay^{2} =
x^{3}.

**Solution:**

The given curve
is ay^{2} = x^{3}

**Maths Class
12 Ex 6.3 Question 21.**

Find the equation
of the normals to the curve y = x^{3} + 2x + 6 which are parallel
to the line x + 14y + 4 = 0.

**Solution:**

Let the required
normal be drawn at the point (x_{1}, y_{1}).

The equation of the given curve is y = x^{3} + 2x + 6 … (i)

**Maths Class
12 Ex 6.3 Question 22.**

Find the
equations of the tangent and normal to the parabola y² = 4ax at the point (at²,
2at).

**Maths Class 12 Ex 6.3 Question 23.**

Prove that the
curves x = y² and xy = k cut at right angles if 8k² = 1.

**Solution:**

The given curves
are x = y² … (i)

And xy = k … (ii)

**Maths Class
12 Ex 6.3 Question 24.**

Find the
equations of the tangent and normal to the hyperbola x^{2}/a^{2}
– y^{2}/b^{2} = 1 at the point (x_{0} ,y_{0}).

**Maths Class
12 Ex 6.3 Question 25.**

Find the equation of the tangent to the curve which is parallel to the line 4x – 2y + 5 = 0.

**Solution:**

Let the point of
contact of the tangent line parallel to the given line be P(x_{1}, y_{1}).

**Choose the
correct answer in Exercises 26 and 27.**

**Maths Class
12 Ex 6.3 Question 26.**

The slope of the
normal to the curve y = 2x² + 3 sin x at x = 0 is

(A) 3

(B) 1/3

(C) –3

(D) –1/3

**Solution:**

(D) The given curve is y = 2x² + 3 sin x

∴
dy/dx = 4x + 3 cos x

At x = 0, dy/dx =
3

∴ Slope of the tangent = 3

⇒ Slope of the normal is –1/3.

Hence,
the correct option is (D).

**Maths Class
12 Ex 6.3 Question 27.**

The line y = x +
1 is a tangent to the curve y² = 4x at the point

(A) (1, 2)

(B) (2, 1)

(C) (1, –2)

(D) (–1, 2)

**Solution:**

(A) The given
curve is y² = 4x

∴ dy/dx = 4/2y = 2/y

Slope of the given line y = x + 1 is 1

∴
2/y = 1 ⇒ y = 2

Putting y = 2 in
y² = 4x, we get 2² = 4x ⇒ x = 1

∴ Point of contact is (1, 2).

Hence, the
correct option is (A).

**NCERT Solutions for Maths Class 12 Exercise 6.1**

**NCERT Solutions for Maths Class 12 Exercise 6.2**