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:

2x² + 7y² = 14

\(\frac {x^2}{7}\) + \(\frac {y^2}{2}\) = 1

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

a² = 7, b² = 2

It passes through (5, 2)

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

2 – 5 m = \(\sqrt { 7m^2 + 2 }\)

(2 – 5 m)² = (7m² + 2)

4 + 25m² – 20m = 7m² + 2

18m² – 20m + 2 = 0

9m² – 10m + 1 = 0

(9m – 1) (m – 1) = 0

m = 1 (or) m = \(\frac {1}{9}\)

Equation of tangent

(i) m= 1 ⇒ (y – 2) = 1(x- 5)

x – y – 3 = 0

(ii) m = \(\frac {1}{9}\) ⇒ y – 2 = \(\frac {1}{9}\) (x – 5)

9y – 18 = x – 5

x – 9y + 13 = 0

Equation of tangent be x – y – 3 = 0 and 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:

Equation of the ellipse

x² + 3y² = 12

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

a² = 12 b² = 4

Equation of the line

x – y + 4 = 0

y = x + 4

m = 1, c = 4

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

c² = a²m² + b²

4² = 12(1)² + 4

16 = 16

∴ The line is a tangent to the ellipse.

Point of contact = (\(\frac {-a^2m}{c}\), \(\frac {b^2}{c}\))

= (\(\frac {-12(1)}{4}\), \(\frac {4}{4}\))

= (-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:

Equation of a tangent be

yt = x + at² Here t = 2

y² = 8x ∴ 4a = 8

a = 2

y(2) = x + 2(4)

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:

Normal equation of the parabola y² = 4ax at t_{1} be

y + xt_{1} = at_{1}³ + 2at_{1}

It meets the point t_{2} {at_{2}², 2at_{2})

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

at_{2}²t_{1} – at_{1}³ = 2at_{1} – 2at_{2}

at_{1}(t_{2}² – t_{1}²) = -2a(t_{2} – t_{1})

at_{1}(t_{2} + t_{1})(t_{2} – t_{1}) = -2a(t_{2} – t_{1})

t_{2} + t_{1} = \(\frac {-2}{t_1}\)

t_{2} = \(\frac {-2}{t_1}\) – t_{1}

t_{2} = -(\(\frac {2}{t_1}\) + t_{1})