## 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,

2y^{2} = 9x …(ii)

From eq^{n}. (i) and (ii),

2y^{2} = 3(-4y – 6)

⇒ 2y^{2} = – 12y – 18

⇒ 2y^{2} + 12y + 18 = 0

⇒ y^{2} ± 6y + 9 = 0

⇒ (y + 3)^{2} = 0 ⇒ y = -3

Putting y = – 3 in eq^{n}. (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 y^{2} = 8x.

Solution:

Equation of parabola,

y^{2} = 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 – x^{2}.

Solution:

Equation of parabola,

y = x – x^{2 }…(i)

Equation of straight line,

x + y = 1 …(ii)

On solving eq. (i) and (ii),

⇒ x + x – x^{2} = 1

⇒ x^{2} – 2x + 1 = 0

This is a quadratic equation, its roots will be equal and coincide

⇒ B^{2} – 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 – x^{2}.

Question 4.

Find the condition that line lx + my + n = 0 touches the parabola y^{2} = 4ax.

Solution:

Equation of parabola,

y^{2} = 4 ax

Equation of straight line,

lx + my + n = 0

⇒ my = -lx – n

Question 5.

Prove that the length of focal chord of parabola y^{2} = 4ax making an angle a with x-axis will be 4a cosec^{2} 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 y^{2} = 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 cosec^{2} α

Thus length of chord PQ = 4a cosec^{2} α.

Question 6.

Find the condition that line x cos α + y sin α – P touches the parabola y^{2} = 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 y^{2} = 4ax to straight line y = mx + c

Question 7.

Find the equation of tangents at the following parabolas:

(i) y^{2} = dx, which is parallel to line 2x – 3y = 4

(ii) y^{2} = 8x, which is perpendicular to line 2x – y + 1 = 0

Solution:

(i) Equation of parabola,

y^{2} = 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 y^{2} = 6x if equation

has equal roots, for which condition,

B^{2} = 4AC

⇒ (4k – 54)^{2} = 4 × 4 × k^{2}

⇒ (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

y^{2} = 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 y^{2} = 8x, if

⇒ 4k^{2} + x^{2} – 4kx – 32x = 0

⇒ x^{2} +(-4k – 32)x + 4k^{2} = 0

Its roots will be same, if

B^{2} = 4AC

⇒ (-4k – 32)^{2} = 4 × 1 × 4k^{2}

⇒ (-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 y^{2} = 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 y^{2} = 8x.

Solution:

From point (x_{1}, y_{1}) two tangents can be drawn at parabola whose combined equation can be find by equation SS’ = T^{2}

Given Point – (4, 0)

Equation of parabola,

⇒ y^{2} = 8x where a = 2

then S = y^{2} – 4ax_{1}

⇒ S = y^{2} – 8x …(i)

S’ = y1^{2} – 4ax^{2}

⇒ S’ = (10)^{2} – 4 × 2 × 4

⇒ S’ = 100 – 32 = 68 …(ii)

T = yy1 – 2a(x + x_{1})

⇒ T = y × 10 – 2 × 2 × (x + 4)

= T = 10y – 4(x + 4)

T^{2} = {10y – 4(x + 4)}^{2}

⇒ T^{2} = 100 y^{2} + 16(x + 4)^{2} – 80(x + 4)y

⇒ T^{2} = 100y^{2} + 16x^{2} + 256 + 128x – 80xy- 320y …(iii)

Put the values from eqn. (i), (ii), (iii) in SS’ – T^{2}

(y^{2} – 8x) (68) = 16x^{2} + 100y^{2} – 80xy + 128x – 320y + 256

⇒ 68y^{2} – 744x – 16x^{2} – 100y^{2}

+ 80xy – 128x+ 320y – 256 = 0

⇒ – 16x^{2} – 32y^{2} + 80xy – 672x + 320y – 256 = 0

⇒ x^{2} + 2y^{2} – 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 y^{2} = 8x

(ii) Upper side of latus rectum of

y^{2} + 12x = 0

Solution:

(i) Equation of parabola,

y^{2} = 8x

Point (2, 4), equation of normal at point (x_{1}, y_{1}) on parabola y^{2} = 4 ax

This is required equation.

(ii) y^{2} + 12x = 0

⇒ y^{2} = -12x

⇒ y^{2} = 4(-3)x

Comparing it by y^{2} = 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) y^{2} = 4x which is parallel to line y – 2x + 5 = 0

(ii) y^{2} = 4x which is perpendicular to line

x + 3y – 1 = 0

Solution:

(i) Equation of parabola,

y^{2} = 4x

Comparing with y^{2} = 4ax

a = 1

Equation of line y – 2x + 5 = 0

⇒ y = 2x – 5

Comparing with y = mx + c

m = 2, c = – 5

At parabola y^{2} = 4ax parallel to line y = mx + c, eq^{n}. of normal from y = mx – 2am – am^{3}

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,

y^{2} = 4x

Comparing with y^{2} = 4ax.

a = 1

Equation of normal of parabola y^{2} = 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 y^{2} = 4ax and its length is 5\(\sqrt { 5a }\)

Solution:

Equation of line,

2x + y – 12a = 0

⇒ y = – 2x + 12a …(i)

Equation of parabola,

y^{2} = 4ax …(ii)

Put value of y from (i) in (ii),

(-2x + 12a)^{2} = 4ax

⇒ 4x^{2} + 144a^{2} + 2(-2x) (12a) – 4ax = 0

⇒ 4x^{2} + 144a^{2} – 48ax – 4ax = 0

⇒ 4x^{2} – 52ax + 144a^{2} = 0

⇒ x^{2} – 13ax + 36a^{2} = 0

⇒ x^{2} – 9ax – 4ax + 36a^{2} = 0

⇒ x(x – 9a) – 4a(x – 9a) = 0

⇒ (x – 9a) (x – 4a) = 0

x = 4a, 9a

Put these value of x in equation (i),

y^{2} = 4a × 4a ⇒ y = 4a

y^{2} = 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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