Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4

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.

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

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
Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4 1
k² = 1024
k = ±32
∴ Equation of tangent
⇒ 10x – 3y ± 32 = 0

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

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)

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

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

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

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
Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4 2
⇒ 4x – 3y = 6
⇒ 4x – 3y – 6 = 0

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

(ii) Equation of the normal to hyperbola be
Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4 3
⇒ 3x + 4y – 42 = 0

Question 7.
Prove that the point of intersection of the tangents at ‘t1‘ and ‘t2‘ on the parabola y² = 4ax is [at1t2, a(t1 + t2)].
Solution:
Equation of the tangent of parabola y² = 4ax be
at t1 yt1 = x + at1² ……….. (1)
at t2 yt2 yt = x + at2² ……….. (2)
(1) – (2) ⇒ y(t1 – t2) = a(t1² – t2²)
y(t1 – t2) = a(t1 + t2)(t1 – t2)
y = a(t1 + t2)
(1) ⇒ t1a(t1 + t2) = x + at1²
x = at1² + at1t2 – at1²
x = at1t2
Point of intersection be [at1t2, a(t1 + t2)]

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

Question 8.
If the normal at the point ‘t1‘ on the parabola y² = 4ax meets the parabola again at the point t2 then prove that t2 = -(t1 + \(\frac {2}{t_1}\))
Solution:
Normal equation of the parabola y² = 4ax at t1 be
y + xt1 = at1³ + 2at1
It meets the point t2 {at2², 2at2)
2 at2 + at2²t1 = at1³ + 2 at1
at2²t1 – at1³ = 2at1 – 2at2
at1(t2² – t1²) = -2a(t2 – t1)
at1(t2 + t1)(t2 – t1) = -2a(t2 – t1)
t2 + t1 = \(\frac {-2}{t_1}\)
t2 = \(\frac {-2}{t_1}\) – t1
t2 = -(\(\frac {2}{t_1}\) + t1)

Samacheer Kalvi 12th Maths Guide Chapter 5 Two Dimensional Analytical Geometry - II Ex 5.4

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