RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.3

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 SP² = PM²

= \(\frac { 1 }{ 17 } \) (h² + 16k² + 9 – 8hk – 24k + 6h)
⇒ 17h² – 68h + 68 + 17k² – 102k + 153 – h² – 16k²
-9 + 8hk + 24k – 6h = 0
⇒ 16h² + 8hk + k² – 74h – 78k + 212 = 0
Thus locus of point P(h. k)
16x² + 8xy + y² – 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 SP² = PM²

Thus, locus of point P(h, k) is y² = 4x + 16, which is required.

Question 2.
Find the vertex, axis, focus and latus rectum for following parabola:
(i) y² = 8x + 8y
(ii) x² + 2y = 8x – 7
Solution:
(i) y² = 8x + 8y
⇒ y² – 8y = 8x
⇒ y² – 2 × 4 × y + 4² = 8x + 4²
⇒ (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
Y² = 8W
⇒ Y² = 4.2.X …(i)
Which is of the form of parbola y² = 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) x² + 2y = 8x – 7
x² – 8x = – 2y – 7
⇒ x² – 2 × 4 × x + 42 = -2y – 7 + 4²
⇒ (x – 4)² = – 2y – 7 + 16
⇒ (x – 4)² = – 2y + 9

In parabola X² = 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 y² = 4ax is 8a. Prove that lines joining the origin to double ordinate will be perpendicular to each other.
Solution:
Equation of parabola,
y² = 4ax
Length of double ordinate = 8a
Let double ordinate is PP’
Given PP’ = 8a
Let OP = x₁ and OP’ = x₁

Then coordinates of P and P’ will be (x₁, 4a) and (x₁, – 4a)
Since point P lies on parabola, y² = 4ax
∴ (4a)² = 4ax₁ ⇒ x₁ = 4a
So, coordinates of P and Q = (4a, 4a) and (4a, – 4a)
∴ m₁ = 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 y²= 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 Y² = 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
y² = 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 ⇒ 9k² = 4ah
Thus, required locus is 9y² = 4ax.

Question 6.
Prove that locus of mid points of all the chords passing through the vertex of parabola y² = 4ax is a parabola y² = 2ax.
Solution:
Given, equation of parabola
y² = 4ax
Let (h, k) be coordinates of mid point P of chord OA passing through vertex O(0, 0) of parabola.
Let (at², 2at) be coordinates of point A.

RBSE Solutions for Class 11 Maths