Math, asked by Ranjan7450, 10 months ago

Find the equation of common tangent to the parabola y^2=4x and x^2=32y

Answers

Answered by stefangonzalez246
6

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

Similar questions