Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4

Question 1.

Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse 2x² + 7y² = 14.

Solution:

2x^{2} + 7y^{2} = 14

(÷ by 14) ⇒ \(\frac{x^{2}}{7}+\frac{y^{2}}{2}\) = 1

comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1

we get a^{2} = 7 and b^{2} = 2

The equation of tangent to the above ellipse will be of the form

y = mx + \(\sqrt{a^{2} m^{2}+b^{2}}\) ⇒ y = mx + \(\sqrt{7 m^{2}+2}\)

Here the tangents are drawn from the point (5, 2)

⇒ 2 = 5m + \(\sqrt{7 m^{2}+2}\) ⇒ 2 – 5m = \(\sqrt{7 m^{2}+2}\)

Squaring on both sides we get

(2 – 5m)^{2} = 7m^{2} + 2

25m^{2} + 4 – 20m – 7m^{2} – 2 = 0

18m^{2} – 20m + 2 = 0

(÷ by 2) ⇒ 9m^{2} – 10m + 1 = 0

(9m – 1) (m – 1) = 0

‘ m = 1 (or) m = 1/9

When m = 1, the equation of tangent is

y = x + 3 or x – y + 3 = 0

When m = \(\frac{1}{9}\) the equation of tangent is 9

y = \(=\frac{x}{9}+\sqrt{\frac{7}{81}+2}\) (i.e.) y = \(\frac{x}{9}+\frac{13}{9}\)

9y = x + 13 or x – 9y + 13 = 0

Question 2.

Find the equations of tangents to the hyperbola \(\frac {x^2}{16}\) – \(\frac {y^2}{64}\) = 1 which are parallel to 10x – 3y + 9 = 0.

Solution:

Equation of Hyperbola \(\frac {x^2}{16}\) – \(\frac {y^2}{64}\) = 1

∴ a² = 16, b² = 64

Tangent is parallel to the line

10x – 3y + 9 = 0 is

10x – 3y + k = 0

∴ 3y = 10x + k

y = \(\frac {10}{3}\)x + \(\frac {k}{3}\)

∴ m = \(\frac {10}{3}\) c = \(\frac {k}{3}\)

Condition that the line y = mx + c to be tangent to the hyperbola is

c² = a²m² – b²

k² = 1024

k = ±32

∴ Equation of tangent

⇒ 10x – 3y ± 32 = 0

Question 3.

Show that the line x – y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12. Also find the co-ordinates of the point of contact.

Solution:

The given ellipse is x^{2} + 3y^{2} = 12

(÷ by 12) ⇒ \(\frac{x^{2}}{12}+\frac{y^{2}}{4}\) = 1

(ie.,) Here a^{2} = 12 and b^{2} = 4

The given line is x – y + 4 = 0

(ie.,) y = x + 4

Comparing this line with y = mx + c

We get m = 1 and c = 4

The condition for the line y = mx + c

To be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is c^{2} = a^{2}m^{2} + b^{2}

LHS = c^{2} = 4^{2} = 16

RHS: a^{2}m^{2} + b^{2} = 12( 1 )^{2} + 4 = 16

LHS = RHS The given line is a tangent to the ellipse. Also the point of contact is

\(\left(\frac{-a^{2} m}{c}, \frac{b^{2}}{c}\right)=\left[-\left(\frac{12(1)}{4}\right), \frac{4}{4}\right]\) (i.e.,) (-3, 1)

Question 4.

Find the equation of the tangent to the parabola y² = 16x perpendicular to 2x + 2y + 3 = 0.

Solution:

Equation of the parabola

y² = 16x

4 a = 16

a = 4

Tangent is perpendicular to the line

2x + 2y + 3 = 0 is 2x – 2y + k = 0

2x – 2y + k = 0

2y = 2x + k

y = x + \(\frac {k}{2}\)

m = 1 c = \(\frac {k}{2}\)

Condition that the line y = mx + c to be tangent to the parabola is

c = \(\frac {a}{m}\)

\(\frac {k}{2}\) = \(\frac {4}{1}\)

k = 8

Equation of the tangent

2x – 2y + 8 = 0

÷ by 2 ⇒ x – y + 4 = 0

Question 5.

Find the equation of the tangent at t = 2 to the parabola y² = 8x (Hint: use parametric form).

Solution:

y^{2} = 8x

Comparing this equation with y^{2} = 4ax

we get 4a = 8 ⇒ a = 2

Now, the parametric form for y^{2} = 4ax is x = at^{2}, y = 2at

Here a = 2 and t = 2

⇒ x = 2(2)^{2} = 8 and y = 2(2) (2) = 8

So the point is (8, 8)

Now eqution of tangent to y^{2} = 4 ax at (x_{1}, y_{1}) is yy_{1} = 2a(x + x_{1})

Here (x_{1}, y_{1}) = (8, 8) and a = 2

So equation of tangent is y(8) = 2(2) (x + 8)

(ie.,) 8y = 4 (x + 8)

(÷ by 4) ⇒ 2y = x + 8 ⇒ x – 2y + 8 = 0

Aliter

The equation of tangent to the parabola y^{2} = 4ax at ‘t’ is

yt = x + at^{2}

Here t = 2 and a = 2

So equation of tangent is

(i.e.,) y(2) = x + 2(2)^{2}

2y = x + 8 ⇒ x – 2y + 8 = 0

Question 6.

Find the equations of the tangent and normal to hyperbola 12x² – 9y² = 108 at θ = \(\frac {π}{3}\) .

(Hint: use parametric form)

Solution:

(i) Equation of the tangent to hyperbola be

⇒ 4x – 3y = 6

⇒ 4x – 3y – 6 = 0

(ii) Equation of the normal to hyperbola be

⇒ 3x + 4y – 42 = 0

Question 7.

Prove that the point of intersection of the tangents at ‘t_{1}‘ and ‘t_{2}‘ on the parabola y² = 4ax is [at_{1}t_{2}, a(t_{1} + t_{2})].

Solution:

Equation of the tangent of parabola y² = 4ax be

at t_{1} yt_{1} = x + at_{1}² ……….. (1)

at t_{2} yt_{2} yt = x + at_{2}² ……….. (2)

(1) – (2) ⇒ y(t_{1} – t_{2}) = a(t_{1}² – t_{2}²)

y(t_{1} – t_{2}) = a(t_{1} + t_{2})(t_{1} – t_{2})

y = a(t_{1} + t_{2})

(1) ⇒ t_{1}a(t_{1} + t_{2}) = x + at_{1}²

x = at_{1}² + at_{1}t_{2} – at_{1}²

x = at_{1}t_{2}

Point of intersection be [at_{1}t_{2}, a(t_{1} + t_{2})]

Question 8.

If the normal at the point ‘t_{1}‘ on the parabola y² = 4ax meets the parabola again at the point t_{2} then prove that t_{2} = -(t_{1} + \(\frac {2}{t_1}\))

Solution:

Equation of normal to y^{2} = 4at’ t’ is y + xt = 2at + at^{3}.

So equation of normal at ‘t_{1}’ is y + xt_{1} = 2at_{1} + at_{1}^{3}

The normal meets the parabola y^{2} = 4ax at ‘t_{2}’ (ie.,) at (at_{2}^{2}, 2at_{2})

⇒ 2at_{2} + at_{1}t_{2}^{2} = 2at_{1} + at_{1}^{3}

So 2a(t_{2} – t_{1}) = at_{1}^{3} – at_{1}t_{2}^{2}

⇒ 2a(t_{2} – t_{1}) = at_{1}(t_{1}^{2} – t_{2}^{2})

⇒ 2(t_{2} – t_{1}) = t_{1}(t_{1} + t_{2})(t_{1} – t_{2})

⇒ 2= -t_{1}(t_{1} + t_{2})

⇒ t_{1} + t_{2} = \(\frac{-2}{t_{1}}\)

⇒ t_{2} = \(-t_{1}-\frac{2}{t_{1}}=-\left(t_{1}+\frac{2}{t_{1}}\right)\)