Find the equation of common tangent to the parabola y^2=4x and x^2=32y
Answers
The equation of common tangent to the parabola y²=4x and x²=32y is x+2y+4=0
Step-by-step explanation:
Given Data
Parabola, y² = 4x and x² = 32 y
for parabola y² = 4ax , then a = 1
Also in x² = 4ay, then a = 8
Equation of tangent is T = 0
ty=x+at2 (value of 'a' is 1) -----> (1)
when t = t1 in y²=4x, the equation of tangent is written as
t1y=x+(t1)2
And at point t2 on x²=32y , equation of Tangent is T=0
t2x=y+8(t2)2 (value of 'a' is 8) ------->(2)
(1) and (2) are the tangent equation of parabolas are similar hence their coefficients(x and y) and constants must be proportional
Tangent equations of the parabolas can be written as
x−t1y+(t1)2=0 -----> (3)
t2x−y−8(t2)2=0 ----->(4)
Equation (3) and (4) represents the line,
1t2=−t1−1=(t1)2−8(t2)2 -----> (5)
By using the condition (5),
take first two terms from (5) we get, (t1)(t2)=1
take last two terms from (5) we get, t1=−(t1)2/(8(t2)2)
from this we get t1=(1÷t2)
then 8t1 = −(t1)4
(t1)3 = −8
t1=−2
On substitute the above conditions the tangent equation for the parabola can be written as, x+2y+4=0
To learn more..
1. https://brainly.in/question/8213023
2. https://brainly.in/question/1170545