## Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.3

Question 1.

Find the equation of parabola whose

(i) Focus (2, 3) and directrix x – 4y + 3 = 0

(ii) Focus (- 3, 0) and directri x + 5 = 0

Solution:

(i) Let P(h, k) is any point on parabola, then According to parabola,

SP = PM or SP^{2} = PM^{2
}

= \(\frac { 1 }{ 17 } \) (h^{2} + 16k^{2} + 9 – 8hk – 24k + 6h)

⇒ 17h^{2} – 68h + 68 + 17k^{2} – 102k + 153 – h^{2} – 16k^{2}

-9 + 8hk + 24k – 6h = 0

⇒ 16h^{2} + 8hk + k^{2} – 74h – 78k + 212 = 0

Thus locus of point P(h. k)

16x^{2} + 8xy + y^{2} – 74x – 78y + 212 = 0, which is required equation of parabola.

(ii) Let P(h, ft) be any variable at Parabola.

According to definition of parabola,

SP = PM or SP^{2} = PM^{2
}

Thus, locus of point P(h, k) is y^{2} = 4x + 16, which is required.

Question 2.

Find the vertex, axis, focus and latus rectum for following parabola:

(i) y^{2} = 8x + 8y

(ii) x^{2} + 2y = 8x – 7

Solution:

(i) y^{2} = 8x + 8y

⇒ y^{2} – 8y = 8x

⇒ y^{2} – 2 × 4 × y + 4^{2} = 8x + 4^{2}

⇒ (y – 4)^{2} = 8x + 16

⇒ (y – 4)^{2} = 8(x + 2) …(i)

For replacing origin at point (4, – 2), put x + 2 and y – 4 = Y

Y^{2} = 8W

⇒ Y^{2} = 4.2.X …(i)

Which is of the form of parbola y^{2} = 4ax where a = 2 and axis X = 0

Coordinates of vertex = (0, 0), coordinates of focus = (0,-1)

Length of Latus Rectum = 4 × 2 = 8

for given parabola (i), put value of X and Y in results.

X = 0 ⇒ x + 2 – 0 ⇒ x = -2

Y = 0 ⇒ y – 4 = 0 ⇒ y = 4

Thus coordinates of vertex = (- 2, 4)

Coordinates of focus

X = 0 ⇒ x + 2 = 2, x = 0

Y = 0 ⇒ y – 4 = 0, y = 4

Thus, coordinates of focus = (0, 4)

Axis Y = 0 ⇒ y – 4 = 0 ⇒ y = 4

Latus rectum = 4a = 4 × 2 = 8

(ii) x^{2} + 2y = 8x – 7

x^{2} – 8x = – 2y – 7

⇒ x^{2} – 2 × 4 × x + 42 = -2y – 7 + 4^{2}

⇒ (x – 4)^{2} = – 2y – 7 + 16

⇒ (x – 4)^{2} = – 2y + 9

In parabola X^{2} = 4ay

Axis X = 0, then x – 4 = 0 ⇒ x = 4

Vertex (0, 0), then

x – 4 = 0 ⇒ x = 4

Question 3.

Length of double ordinate of parabola y^{2} = 4ax is 8a. Prove that lines joining the origin to double ordinate will be perpendicular to each other.

Solution:

Equation of parabola,

y^{2} = 4ax

Length of double ordinate = 8a

Let double ordinate is PP’

Given PP’ = 8a

Let OP = x_{1} and OP’ = x_{1
}

Then coordinates of P and P’ will be (x_{1}, 4a) and (x_{1}, – 4a)

Since point P lies on parabola, y^{2} = 4ax

∴ (4a)^{2} = 4ax_{1} ⇒ x_{1} = 4a

So, coordinates of P and Q = (4a, 4a) and (4a, – 4a)

∴ m_{1} = Slope of line OP

Question 4.

If vertex and focus of parabola are at a distance of a and a’ from origin to x-axis, then prove that equation of parabola will be y^{2}= 4(a’ – a)(x – a).

Solution:

According to question, coordinates of x-axis of vertex of parbola is at a distance a.

Thus vertex of parabola will be (a, 0) and focus (a’, 0). Vertex of parabola (a, 0) and length of latus rectum

= 4(Distance between focus and vertex)

= 4(a’ – a)

Since parabola is symmetric to x – axis.

Then from equation of parabola Y^{2} = 4AX

y = 4 × (a’ – a) × (x – a)

⇒ y = 4(a’ – a) (x – a)

This is required equation.

Question 5.

PQ is double ordinate of a parabola. Find the locus of its trisection point.

Solution:

Given, equation of parabola

y^{2} = 4ax

Let R, S trisects double ordinate PQ.

Let double ordinate PQ meets A-axis at point A.

Let coordinates of point R are (h, k)

then OA – h, AR = k

∴ RS = RA + AS = k + k – 2k

⇒ PR = RS = SQ = 2k

⇒ PA =RA+PR = k + 2k = 3k

∴ Coordinates of point P are (h, 3k),

∵ Point P lies on parabola

∴ (3k)^{2} = 4ah ⇒ 9k^{2} = 4ah

Thus, required locus is 9y^{2} = 4ax.

Question 6.

Prove that locus of mid points of all the chords passing through the vertex of parabola y^{2} = 4ax is a parabola y^{2} = 2ax.

Solution:

Given, equation of parabola

y^{2} = 4ax

Let (h, k) be coordinates of mid point P of chord OA passing through vertex O(0, 0) of parabola.

Let (at^{2}, 2at) be coordinates of point A.

## Leave a Reply