Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.4
Question 1.
Find the coordinates of the points of intersection where straight line 4y + 3x + 6 = 0 cuts the parabola 2y2 = 9x.
Solution:
Equation of straight line,
4y + 3x + 6 = 0
⇒ 3x = -4y – 6 …(i)
Equation of parabola,
2y2 = 9x …(ii)
From eqn. (i) and (ii),
2y2 = 3(-4y – 6)
⇒ 2y2 = – 12y – 18
⇒ 2y2 + 12y + 18 = 0
⇒ y2 ± 6y + 9 = 0
⇒ (y + 3)2 = 0 ⇒ y = -3
Putting y = – 3 in eqn. (ii),
9x = 2 × (-3)2 = 2 × 9
⇒ x = 2
Thus coordinates of intersection point of (2, – 3).
Question 2.
Find the length of chord cut by line 4y – 8 at parabola y2 = 8x.
Solution:
Equation of parabola,
y2 = 8x …(i)
Equation of line,
4y – 3x = 8
⇒ 4y = 3x + 8
Thus, coordinates of ends of chord are (8,8) and (\(\frac { 8 }{ 9 } \),\(\frac { 8 }{ 3 } \)) Thus, length of chord
Thus, length of chord is \(\frac { 80 }{ 9 } \) units.
Question 3.
Prove that straight line x + y = 1 touches parabola y = x – x2.
Solution:
Equation of parabola,
y = x – x2 …(i)
Equation of straight line,
x + y = 1 …(ii)
On solving eq. (i) and (ii),
⇒ x + x – x2 = 1
⇒ x2 – 2x + 1 = 0
This is a quadratic equation, its roots will be equal and coincide
⇒ B2 – 4AC = 0
⇒ (- 2)2 – 4(1) (1) = 0
⇒ 4 – 4 = 0
⇒ 0 = 0
Hence, straight line x + y = 1 will touch the parabola
y = x – x2.
Question 4.
Find the condition that line lx + my + n = 0 touches the parabola y2 = 4ax.
Solution:
Equation of parabola,
y2 = 4 ax
Equation of straight line,
lx + my + n = 0
⇒ my = -lx – n
Question 5.
Prove that the length of focal chord of parabola y2 = 4ax making an angle a with x-axis will be 4a cosec2 a.
Solution:
coordinates of ends P and Q of chord PQ, making an angle α with x-axis and passing through focus S (a, 0) of parabola y2 = 4 ax.
From point Q, draw a line parallel to x – axis and draw PR ⊥ from point P to that line.
PQ – 2a . 2 cosec α . cosec α
= 4a cosec2 α
Thus length of chord PQ = 4a cosec2 α.
Question 6.
Find the condition that line x cos α + y sin α – P touches the parabola y2 = 4ax
Solution:
Equation of straight line,
x cos α + y sin α = p
⇒ y sin α = – x cos α + p
By the condition of tangency of parabola y2 = 4ax to straight line y = mx + c
Question 7.
Find the equation of tangents at the following parabolas:
(i) y2 = dx, which is parallel to line 2x – 3y = 4
(ii) y2 = 8x, which is perpendicular to line 2x – y + 1 = 0
Solution:
(i) Equation of parabola,
y2 = 6x
Equation of straight line,
2x – 3y = 4
⇒ 2x – 3y – 4 = 0
Equation of line parallel to this line,
2x – 3y + k = 0
Line (i) will touch parabola y2 = 6x if equation
has equal roots, for which condition,
B2 = 4AC
⇒ (4k – 54)2 = 4 × 4 × k2
⇒ (4k – 54)2 – (4k)2 = 0
⇒ (4k – 54 – 4k) (4k – 54 + 4k) = 0
⇒ (-54) (8k – 54) = 0
This is required equation.
(ii) Equation of parabola
y2 = 8x
Equation of line,
2x – y + 1 = 0
Equation of line perpendicular to this line,
x + 2y – 2k = 0 …(i)
Line (i) will touch the parabola y2 = 8x, if
⇒ 4k2 + x2 – 4kx – 32x = 0
⇒ x2 +(-4k – 32)x + 4k2 = 0
Its roots will be same, if
B2 = 4AC
⇒ (-4k – 32)2 = 4 × 1 × 4k2
⇒ (-4k – 32)2 – (4k)2 = 0
⇒ (- 4k – 32 – 4k)(-4k – 32 + 4k) = 0
⇒ 8k = -32
⇒ k = -4
Putting value of k in equation (i)
x + 2y – 4 = 0
This is required equation.
Question 8.
For which value of k line 2x – 3y-k will touch parabola y2 = 6x ?
Solution:
Equation of line,
2x – 3y = k
⇒ 3y = 2x – k
Question 9.
Find the equation of tangents which are drawn from point (4, 10) to parabola y2 = 8x.
Solution:
From point (x1, y1) two tangents can be drawn at parabola whose combined equation can be find by equation SS’ = T2
Given Point – (4, 0)
Equation of parabola,
⇒ y2 = 8x where a = 2
then S = y2 – 4ax1
⇒ S = y2 – 8x …(i)
S’ = y12 – 4ax2
⇒ S’ = (10)2 – 4 × 2 × 4
⇒ S’ = 100 – 32 = 68 …(ii)
T = yy1 – 2a(x + x1)
⇒ T = y × 10 – 2 × 2 × (x + 4)
= T = 10y – 4(x + 4)
T2 = {10y – 4(x + 4)}2
⇒ T2 = 100 y2 + 16(x + 4)2 – 80(x + 4)y
⇒ T2 = 100y2 + 16x2 + 256 + 128x – 80xy- 320y …(iii)
Put the values from eqn. (i), (ii), (iii) in SS’ – T2
(y2 – 8x) (68) = 16x2 + 100y2 – 80xy + 128x – 320y + 256
⇒ 68y2 – 744x – 16x2 – 100y2
+ 80xy – 128x+ 320y – 256 = 0
⇒ – 16x2 – 32y2 + 80xy – 672x + 320y – 256 = 0
⇒ x2 + 2y2 – 5xy + 42x – 20y + 16 = 0
This is required equation.
Question 10.
Find the equation of normal at the following parabolas:
(i) At point (2, 4) on y2 = 8x
(ii) Upper side of latus rectum of
y2 + 12x = 0
Solution:
(i) Equation of parabola,
y2 = 8x
Point (2, 4), equation of normal at point (x1, y1) on parabola y2 = 4 ax
This is required equation.
(ii) y2 + 12x = 0
⇒ y2 = -12x
⇒ y2 = 4(-3)x
Comparing it by y2 = 4ax
a = – 3
Equation of latus rectum x = a
⇒ x = – 3
Putting x = – 3 in the equation of parabola
Question 11.
Find the equation of normal at following parabolas :
(i) y2 = 4x which is parallel to line y – 2x + 5 = 0
(ii) y2 = 4x which is perpendicular to line
x + 3y – 1 = 0
Solution:
(i) Equation of parabola,
y2 = 4x
Comparing with y2 = 4ax
a = 1
Equation of line y – 2x + 5 = 0
⇒ y = 2x – 5
Comparing with y = mx + c
m = 2, c = – 5
At parabola y2 = 4ax parallel to line y = mx + c, eqn. of normal from y = mx – 2am – am3
y = 2x – 2 × 1 × 2 – 1 × (2)3
⇒ y = 2x – 4 – 8
⇒ 7 – 2x + 12 = 0
⇒ 2x – y – 12 =0
This is required equation.
(ii) Equation of parabola,
y2 = 4x
Comparing with y2 = 4ax.
a = 1
Equation of normal of parabola y2 = 4ax, perpendicular to line 7 = mx + c
⇒ y = 3x – 2 × 1 × 3 – 1 × (3)3
⇒ y = 3x – 6 – 27
⇒ y = 3x – 33
⇒ 3x – y – 33 = 0
This is required equation.
Question 12.
Prove that line 2x + y – 12a = 0 is normal chord at parabola y2 = 4ax and its length is 5\(\sqrt { 5a }\)
Solution:
Equation of line,
2x + y – 12a = 0
⇒ y = – 2x + 12a …(i)
Equation of parabola,
y2 = 4ax …(ii)
Put value of y from (i) in (ii),
(-2x + 12a)2 = 4ax
⇒ 4x2 + 144a2 + 2(-2x) (12a) – 4ax = 0
⇒ 4x2 + 144a2 – 48ax – 4ax = 0
⇒ 4x2 – 52ax + 144a2 = 0
⇒ x2 – 13ax + 36a2 = 0
⇒ x2 – 9ax – 4ax + 36a2 = 0
⇒ x(x – 9a) – 4a(x – 9a) = 0
⇒ (x – 9a) (x – 4a) = 0
x = 4a, 9a
Put these value of x in equation (i),
y2 = 4a × 4a ⇒ y = 4a
y2 = 4a × 9a ⇒ y = 6a
Thus coordinates of ends of latus rectum are (4a, 4a) and (9a, – 6a)
∵ Coordinates will lie in 1st and 4th quadrant will be (+, +) and (+, -).
Thus length of latus rectum
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