The equation of a common tangent to the parab | vedclass

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

Question

(B) Let the equation of the tangent to the parabola $y = x^{2}$ be $y = mx - \frac{m^{2}}{4}$.Since this line is also tangent to $y = -(x - 2)^{2}$,we

Vedclass · Accepted Answer

$y = 4(x - 1)$

Vedclass · Answer

(B) Let the equation of the tangent to the parabola $y = x^{2}$ be $y = mx - \frac{m^{2}}{4}$.Since this line is also tangent to $y = -(x - 2)^{2}$,we substitute $y$ in the second equation:$mx - \frac{m^{2}}{4} = -(x - 2)^{2}$$mx - \frac{m^{2}}{4} = -(x^{2} - 4x + 4)$$x^{2} + x(m - 4) + 4 - \frac{m^{2}}{4} = 0$For the line to be a tangent,the discriminant $D$ must be zero:$D = (m - 4)^{2} - 4(1)(4 - \frac{m^{2}}{4}) = 0$$m^{2} - 8m + 16 - 16 + m^{2} = 0$$2m^{2} - 8m = 0$$2m(m - 4) = 0$Thus,$m = 0$ or $m = 4$.For $m = 4$,the tangent equation is $y = 4x - \frac{4^{2}}{4} = 4x - 4 = 4(x - 1)$.

The equation of a common tangent to the parabolas $y = x^{2}$ and $y = -(x - 2)^{2}$ is:

Similar Questions

If the straight line $x + y = 1$ touches the parabola $y^2 - y + x = 0$,then the coordinates of the point of contact are

The intercepts of the focal chord (which is a part of the latus rectum) to the parabola $y^{2}=4ax$ are $b$ and $k$. Then $k$ is equal to:

The number of common tangents to the parabolas $y = x^{2}$ and $y = -x^{2} + 4x - 4$ is:

If two tangents drawn from a point $P$ to the parabola $y^2 = 4x$ are such that the slope of one tangent is double the other,then $P$ lies on the curve:

The tangent to the parabola $y^2 = 4x$ at the point where it intersects the circle $x^2 + y^2 = 5$ in the first quadrant,passes through the point

Vedclass Test Series

Exam Paper Generator

Online Exam Module