Answer :
We have y^2 = 20x .
Let (x1,y1) be the tangential point
Now 2yy′ = 20
∴ y′ = 10/y ie., at (x1, y1) m = 10/y1 …(1)
But the tangent makes an angle 45° with the x – axis.
∴ slope of the tangent m=tan 45° = 1 … (2)
From (1) and (2)
10/y1 = 1 ⇒ y1 = 10
But (x1,y1) lies on y^2 = 20x ⇒ y1^2 = 20 x1
100 = 20(x1) or x1 = 5
i.e., (x1,y1) = (5,10) and hence the equation of the tangent at (5, 10) is
y – 10 = 1(x – 5)
y = x + 5
Let (x1,y1) be the tangential point
Now 2yy′ = 20
∴ y′ = 10/y ie., at (x1, y1) m = 10/y1 …(1)
But the tangent makes an angle 45° with the x – axis.
∴ slope of the tangent m=tan 45° = 1 … (2)
From (1) and (2)
10/y1 = 1 ⇒ y1 = 10
But (x1,y1) lies on y^2 = 20x ⇒ y1^2 = 20 x1
100 = 20(x1) or x1 = 5
i.e., (x1,y1) = (5,10) and hence the equation of the tangent at (5, 10) is
y – 10 = 1(x – 5)
y = x + 5