The line y=x+1 is a tangent to the curve y^{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the given curve is $y^{2}=4x$. Differentiating with respect to $x$,we have: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \fra

Vedclass · Accepted Answer

$(1,2)$

Vedclass · Answer

(A) The equation of the given curve is $y^{2}=4x$. Differentiating with respect to $x$,we have: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$. 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 $y=mx+c$. The slope of the 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. Thus,we must have $\frac{2}{y} = 1 \Rightarrow y=2$. Substituting $y=2$ into the equation of the line $y=x+1$,we get $2 = x+1 \Rightarrow x=1$. Hence,the line $y=x+1$ is a tangent to the given curve at the point $(1, 2)$. The correct answer is $A$.

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

Similar Questions

If the line $lx + my + n = 0$ is tangent to the parabola $y^{2} = 4ax$,then

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$ which is at minimum distance from the circle $x^2 + y^2 - 4x - 20y + 103 = 0$. Then $\alpha \beta$ is

Suppose $AB$ is a focal chord of the parabola $y^2=12x$ of length $l$ and slope $m < \sqrt{3}$. If the distance of the chord $AB$ from the origin is $d$,then $l \cdot d^2$ is equal to ....................

The slope of the line touching both the parabolas $y^2 = 4x$ and $x^2 = -32y$ is

The line $y = mx + 1$ is a tangent to the parabola $y^2 = 4x$. Find the value of $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module