Advertisements
Advertisements
प्रश्न
Prove that the line 2x - 3y = 9 touches the conics y2 = -8x. Also, find the point of contact.
उत्तर
y2 = - 8x ...(i)
y2 = - 4ax
⇒ a = 2
2x - 3y = 9 ...(ii)
⇒ `(2"x")/3 - 9/3 = "y"` ⇒ y = `2/3` x - 3
m = `2/3` , c = -3
Substituting (ii) in (i) we get :
⇒ `((2"x"-9)/3)^2` = - 8x
⇒ `(4"x"^2 + 81 -36"x")/9 = -8"x"`
⇒ 4x2 + 81 - 36x = -72x
⇒ 4x2 - 36x + 72x +81 = 0
⇒ 4x2 + 36x + 81 = 0
⇒ x = `(-36 ± sqrt((36)^2 - 4xx4xx81))/8`
x = `(-36 ± 0)/8`
x = `(-36)/8 = -9/2`
⇒ x = `-9/2, -9/2`
So the line (ii) meet the parabola (i) at two co-incident points, hence line (ii) meet the parabola (i)
y = `2/3x - 3 = 2/3((-9)/2) -3`
= - 3 - 3 = -6
∴ Point of contact is `(-9/2,-6)`
APPEARS IN
संबंधित प्रश्न
Let f : N→N be a function defined as f(x)=`9x^2`+6x−5. Show that f : N→S, where S is the range of f, is invertible. Find the inverse of f and hence find `f^-1`(43) and` f^−1`(163).
If the function f : R → R be given by f[x] = x2 + 2 and g : R → R be given by `g(x)=x/(x−1)` , x≠1, find fog and gof and hence find fog (2) and gof (−3).
Consider `f:R - {-4/3} -> R - {4/3}` given by f(x) = `(4x + 3)/(3x + 4)`. Show that f is bijective. Find the inverse of f and hence find `f^(-1) (0)` and X such that `f^(-1) (x) = 2`
Let f : N → ℝ be a function defined as f(x) = 4x2 + 12x + 15. Show that f : N → S, where S is the range of f, is invertible. Also find the inverse of f.
Let R be the equivalence relation in the set Z of integers given by R = {(a, b): 2 divides a – b}. Write the equivalence class [0]
If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g
If f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, write the range of f and g
If A = {1, 2, 3} and f, g are relations corresponding to the subset of A × A indicated against them, which of f, g is a function? Why?
f = {(1, 3), (2, 3), (3, 2)}
g = {(1, 2), (1, 3), (3, 1)}
Let f: R → R be defined by f(x) = sin x and g: R → R be defined by g(x) = x 2 , then f o g is ______.
Let f, g: R → R be defined by f(x) = 2x + 1 and g(x) = x2 – 2, ∀ x ∈ R, respectively. Then, find gof
If functions f: A → B and g: B → A satisfy gof = IA, then show that f is one-one and g is onto
Functions f , g: R → R are defined, respectively, by f(x) = x 2 + 3x + 1, g(x) = 2x – 3, find f o g
Functions f , g: R → R are defined, respectively, by f(x) = x 2 + 3x + 1, g(x) = 2x – 3, find g o f
Functions f , g: R → R are defined, respectively, by f(x) = x 2 + 3x + 1, g(x) = 2x – 3, find f o f
Functions f , g: R → R are defined, respectively, by f(x) = x 2 + 3x + 1, g(x) = 2x – 3, find g o g
Let f: R → R be defined by f(x) = `{{:(2x",", x > 3),(x^2",", 1 < x ≤ 3),(3x",", x ≤ 1):}`. Then f(–1) + f(2) + f(4) is ______.
If f : R → R and g : R → R defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the value of x for which f(g(x)) = 25 is ____________.
Let f : R → R be given by f(x) = tan x. Then f-1(1) is ____________.
Let f(x) = x2 – x + 1, x ≥ `1/2`, then the solution of the equation f(x) = f-1(x) is ____________.
