Find the points on the curve frac{x^{2}}{4} + | vedclass

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

Question

(A) Given the equation of the curve: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$. Differentiating both sides with respect to $x$,we get: $\frac{2x}{4} +

Vedclass · Accepted Answer

$(0, 5)$ and $(0, -5)$

Vedclass · Answer

(A) Given the equation of the curve: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$. Differentiating both sides with respect to $x$,we get: $\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0$ $\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{25x}{4y}$. The tangent is parallel to the $x$-axis if the slope $\frac{dy}{dx} = 0$. Setting the slope to zero: $-\frac{25x}{4y} = 0$,which implies $x = 0$. Substituting $x = 0$ into the original equation of the curve: $\frac{0^{2}}{4} + \frac{y^{2}}{25} = 1$ $\frac{y^{2}}{25} = 1$ $y^{2} = 25$ $y = \pm 5$. Thus,the points at which the tangents are parallel to the $x$-axis are $(0, 5)$ and $(0, -5)$.

Find the points on the curve $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ at which the tangents are parallel to the $x$-axis.

Similar Questions

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

If $y=4x-5$ is tangent to the curve $y^2=px^3+q$ at $(2,3)$,then

The line which is parallel to the $x$-axis and intersects the curve $y = \sqrt{x}$ at an angle of $45^\circ$ is given by:

The angle between the curves $y^2=2x$ and $x^2+y^2=8$ is

Tangents are drawn to the curve $y = \sin x$ from the origin. The locus of the points of contact is

Vedclass Test Series

Exam Paper Generator

Online Exam Module