The number of points on the curve y=2t^2+3t-5 | vedclass

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

Question

(D) The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t+3}{3t^2-8t-3}$.Since the normal is parallel to t

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t+3}{3t^2-8t-3}$.Since the normal is parallel to the $X$-axis,the tangent at that point must be perpendicular to the $X$-axis,which means the tangent is parallel to the $Y$-axis.Thus,the slope of the tangent $\frac{dy}{dx}$ must be undefined,or $\frac{dx}{dt} = 0$.Setting the denominator to zero: $3t^2-8t-3 = 0$.Solving the quadratic equation: $(3t+1)(t-3) = 0$.This gives $t = 3$ and $t = -\frac{1}{3}$.For these values of $t$,the slope of the tangent is undefined,implying the normal is parallel to the $X$-axis.Therefore,there are $2$ such points.

The number of points on the curve $y=2t^2+3t-5$ and $x=t^3-4t^2-3t$ such that the normals drawn at them on the curve are parallel to the $X$-axis is

Similar Questions

If $\theta$ is the angle between the curves $xy=2$ and $x^2+4y=0$,then $\tan \theta$ is equal to:

The sum of the intercepts on the axes cut off by the tangent to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$ is:

The equation of the normal to the curve $2x^{2} + 3y^{2} - 5 = 0$ at the point $P(1, 1)$ is:

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

Find the equation of the tangent to the curve $y = be^{-x/a}$ at the point where it crosses the $y$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module