Question

if y=x+a is tanget to parabola y^2=4(x+1) then value of a ?

Answer 1

y = x + a has a slope of 1. So let's take the derivative of y^2 and find where it's a:\[2y \frac{ dy }{ dx }=4 \implies \frac{ dy }{ dx } = \frac{ 2 }{ y }\]\[\frac{ 2 }{ y } = 1\implies y = 2\] So now we know that the tangent line occurs at point (x, 2). Let's now find x by replacing y with 2 in original equation and solving for x:\[(2)^2=4x+4 \implies 4x = 0 \implies x = 0\] So the slope of the tangent is 1 and it occurs at point (0, 2). Now we can solve for a:\[y = x + a \rightarrow 2 = 0 + a \implies a = 2\] Therefore:\[y=x+2\]\[a=2\]
@shiv99