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.
