What's the equation of a tangent on the parabola
y2=5x at the point (5, 5)
x - 2y + 5 = 0
Method 1
Given equation of parabola is,
y2=5x .............(1)
The tangent will be having a slope equal to dydx∣∣(5,5)
and also passes through the point (5,5)
∴ differentiating (1) with respect to x
2y.dydx=5
dydx=52y∣∣(5,5)=12
∴ equation of the tangent
y−5=12(x−5)
2y-10=x-5
x-2y+5=0
Method 2
Here we can use the formula T1=0 for tangent
at the point (x1,y1).T1 is obtained by replacing
−x2 with xx1
−y2 with yy1
−x with x+x12
−y with y+y12
∴ T1=0
⇒ yy1=5.[x+x12]
⇒ 5y=5.(x+5)2
⇒2y=x+5
⇒x−2y+5=0