Question

Chose the correct answer.

The line y=x+1 is a tangent to the curve $y^2=4x$ at the point

(a) (1,2) (b) (2,1) (c) (1,-2) (d) (-1,2).

Answer 1

The equation of the given curve is $y^2=4x$ ....(i)

On differentiating w.r.t.x, we get

Therefore, the slope of the tangent to the given curve at any point (x, y) is given by

$\frac{dy}{dx}=\frac{2}{y}$

The given line is y=x+1(which is of the form y=mx+c)

$\therefore$ Slope of this line is 1.

The line y=x+1 is a tangent to the given curve, if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.

Thus, we must have $\frac{2}{y}=1\Rightarrow =2$

On putting y=2 in Eq. (i), we get $2^2=4x\Rightarrow x=1$

Hence, the line y=x+1 is a tangent to the given curve at the point (1,2). So, the correct option is (a).