ParabolaHard
Question
The curve described parametrically by x = t2 + t +1, y = t2 - t +1 represents
Options
A.a pair of straght lines
B.an ellipse
C.a parabola
D.a hyperbola
Solution
Given curves are x = t2 + t + 1 ......(i)
and y = t2 - t + 1 ......(ii)
On subtracting Eq. (ii) from Eq. (ii),
x - y = 2t
Thus, x = t2 + t + 1
⇒
⇒ 4x = (x - y)2 + 2x - 2y + 4
⇒ (x - y)2 = 2(x + y - 2)
⇒ x2 + y2 - 2xy - 2x - 2y + 4 = 0
Now, ᐃ = 1.1.4 + 2.(-1)(-1)(-1)
-1 ×(-1)2 -1 ×(-1)2 - 4(-1)2
= 4 - 2 -1-1- 4 = -4
∴ ᐃ ≠ 0
and ab - h2 = 1.1 - (-1)2 = 1 - 1 = 0
Hence, it represents a rquation of parabola.
and y = t2 - t + 1 ......(ii)
On subtracting Eq. (ii) from Eq. (ii),
x - y = 2t
Thus, x = t2 + t + 1
⇒

⇒ 4x = (x - y)2 + 2x - 2y + 4
⇒ (x - y)2 = 2(x + y - 2)
⇒ x2 + y2 - 2xy - 2x - 2y + 4 = 0
Now, ᐃ = 1.1.4 + 2.(-1)(-1)(-1)
-1 ×(-1)2 -1 ×(-1)2 - 4(-1)2
= 4 - 2 -1-1- 4 = -4
∴ ᐃ ≠ 0
and ab - h2 = 1.1 - (-1)2 = 1 - 1 = 0
Hence, it represents a rquation of parabola.
Create a free account to view solution
View Solution FreeMore Parabola Questions
If the focus of the parabola (y - λ)2 = 4(x - λ) always lies between the lines 2x + y = 1 and 2x + y = 3 then ...F1 and F2 are focus of hyperbola 16x2 - 9y2 - 32x = 128.S = 0 is equation of parabola whose vertex is F1 and focus is F2...AB, AC are tangents to a parabola y2 = 4ax, p1 p2 and p3 are the lengths of the perpendiculars from A, B and C respectiv...If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y2 = 4ax and x...Let one end of a focal chord of the parabola $y^2 = 16x$ be $A(16, 16)$. If $P(\alpha, \beta)$ divides this focal chord ...