The point on the curve x^2+y^2=a^2, y geq 0,a | vedclass

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

Question

(C) Given the curve equation: $x^2+y^2=a^2$. Differentiating both sides with respect to $x$: $2x + 2y \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{x}{y}

Vedclass · Accepted Answer

$(0, a)$

Vedclass · Answer

(C) Given the curve equation: $x^2+y^2=a^2$. Differentiating both sides with respect to $x$: $2x + 2y \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{x}{y}$. Since the tangent is parallel to the $x$-axis,the slope of the tangent must be zero: $\frac{dy}{dx} = 0$ $-\frac{x}{y} = 0 \implies x = 0$. Substituting $x = 0$ into the curve equation: $0^2 + y^2 = a^2 \implies y^2 = a^2$. Since $y \geq 0$,we get $y = a$. Thus,the point on the curve is $(0, a)$.

The point on the curve $x^2+y^2=a^2, y \geq 0$,at which the tangent is parallel to the $x$-axis is

Similar Questions

If the length of the tangent drawn from the point $(5, 3)$ to the circle $x^2 + y^2 + 2x + ky + 17 = 0$ is $7$,then the value of $k$ is:

$A$ circle touches the $y$-axis at the point $(0,4)$ and passes through the point $(2,0)$. Which of the following lines is not a tangent to this circle?

The length of the tangent from the point $(4, 5)$ to the circle $x^2 + y^2 + 2x - 6y - 6 = 0$ is

If $O$ is the origin and $OP$,$OQ$ are tangents to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,the circumcentre of the triangle $OPQ$ is

The circle $x^2+y^2-4x-8y+16=0$ rolls along the tangent drawn to it at $(2+\sqrt{3}, 3)$ by $2$ units. The equation of the circle in the new position is

Vedclass Test Series

Exam Paper Generator

Online Exam Module