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

## 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:
$$\frac{x^{2}}{16}-\frac{y^{2}}{64}$$ = 1
Here a2 = 16 and b2 = 64
The equation of tangents will be of the form y = mx ± $$\sqrt{a^{2} m^{2}-b^{2}}$$
(i.e.,) y = mx ± $$\sqrt{16 m^{2}-64}$$
Where ‘m’ is the slope of the tangent.
Here we are given that the tangents are parallel to 10x – 3y + 9 = 0
So the slope of tangents will be equal to the slope of the given line

Question 3.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x2 + 3y2 = 12. Also find the coordinates 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:
y2 =16x
Comparing this equation with y2 = 4ax
we get 4a = 16 ⇒ a = 4
The equation of tangent to the parabola y2 = 16x will be of the form

So m = 1
⇒ The equation of tangent will be y = 1(x) + $$\frac{4}{1}$$ (i.e.,) y = x + 4
(or) 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 = $$\frac{\pi}{3}$$ .
(Hint: Use parametric form)
Solution:
12x2 – 9y2 = 108

Normal is a line perpendicular to tangent
So equation of normal will be of the form 3x + 4y + k = 0
The normal is drawn at (6, 6)
⇒ 18 + 24 + k = 0 ⇒ k = – 42
So equation of normal is 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 [at1 t2, a (t1 + t2)].
Solution:
The equation of tangent to y2 = 4ax at ‘t’ is given by yt = x + at2
So equation of tangent at ‘t1‘ is yt1 = x + at12
and equation of tangent at ‘t2‘ is yt2 = x + at22
To find the point of intersection we have to solve the two equations

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 = $$-\left(t_{1}+\frac{2}{t_{1}}\right)$$
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 Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.4 Additional Questions

Question 1.
Find the equations of the tangent and normal to the parabolas: x2 + 2x – 4y + 4 = 0 at (0, 1)
Solution:

Question 2.
Find the equations of the tangent and normal to the parabola y2 = 8x at t = $$\frac{1}{2}$$
Solution:

∴ Equation of the tangent is 2x – y + 1 = 0 and equation of the normal is 2x + 4y – 9 = 0

Question 3.
Find the equations of the tangents: to the parabola y2 = 6x, parallel to 3x – 2y + 5 = 0
Solution:

∴ Equation of the tangent is 3x – 2y + 2 = 0

Question 4.
Find the equations of the tangents: to the parabola 4x2 – y2 = 64 Which are parallel to 10x – 3y + 9 = 0.
Solution:

Question 5.
Find the equation of the tangent from point (2, -3) to the parabola y2 = 4x.
Solution:
y2 = 4x
Equation of the tangent to the parabola will be of the form y = $$m x+\frac{1}{m}$$
The tangents pass through (2, -3)

The two tangents drawn from (2, -3) are x + y + 1 = 0, x + 2y + 4 = 0

Question 6.
Find the equation of the tangents from the point (2, -3) to the parabola 2x2 – 3y2 = 6
Solution:
2x2 – 3y2 = 6

Question 7.
Prove that the line 5x + 12y = 9 touches the hyperbola x2 – 9y2 = 9 and find the point of contact.
Solution:

(1) = (2) ⇒ The given line is a tangent to the curve i.e., the given line touches the curve. To find the point of contact we have to solve the line and the curve.
The given line 5x + 12y = 9

Question 8.
Show that the line x – y + 4 = 0 is a tangent to the ellipse x2 + 3y2 = 12. Find the co-ordinates of the point of contact.
Solution:
The given ellipse is x2 + 3y2 = 12
Given line is x – y + 4 = 0 ⇒ y = x + 4
Here, m = 1 and c = 4
The condition for the line y = mx + c to be a tangent to the ellipse

RHS: a2 m2 + b = 12(1) + 4 = 16
LHS: RHS ⇒ the given lines is a tangent to the ellipse.

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