find the equation of tangent to the parabola y²=9x at (1,-3)
Answers
EXPLANATION.
Equation of tangent to the parabola,
⇒ y² = 9x at (1,-3).
As we know that,
General equation of parabola y² = 4ax.
Compare both the equation, we get.
⇒ 4a = 9.
⇒ a = 9/4.
As we know that,
Slope form : The equation of the tangent of the parabola,
⇒ y² = 4ax is y = mx + a/m.
Put the value in the equation, we get.
⇒ y = mx + 9/4m.
Put the value of x & y in equation, we get.
⇒ (-3) = m(1) + 9/4m.
⇒ -3 = 4m² + 9/4m.
⇒ -12m = 4m² + 9.
⇒ 4m² + 12m + 9 = 0.
Factorizes the equation into middle term splits, we get.
⇒ 4m² + 6m + 6m + 9 = 0.
⇒ 2m(2m + 3) + 3(2m + 3) = 0.
⇒ (2m + 3)(2m + 3) = 0.
⇒ (2m + 3)² = 0.
⇒ (2m + 3) = 0.
⇒ 2m = -3.
⇒ m = -3/2.
As we know that,
Equation of the tangent.
⇒ y - y₁ = m(x - x₁).
Put the value in the equation, we get.
⇒ (y - (-3)) = -3/2(x - 1).
⇒ (y + 3) = -3/2(x - 1).
⇒ 2(y + 3) = -3(x - 1).
⇒ 2y + 6 = -3x + 3.
⇒ 2y + 3x - 3 = 0.
MORE INFORMATION.
Equation of normal.
(1) = Point form : The equation to the normal at the point (x₁, y₁) of the parabola y² = 4ax is given by y - y₁ = m(x - x₁).
(2) = Parametric form : The equation to the normal at the point (at², 2at) is y + tx = 2at + at³.
(3) = Slope form : Equation of normal in terms of slope m is y = mx - 2am - am³.
(4) = The foot of the normal is (am², -2am).
Condition of Normal.
The line y = mx + c is a normal to the parabola.
(1) = y² = 4ax if c = -2am - am³.
(2) = x² = 4ay if c = 2a + a/m².