If x-2y+k=0 is a tangent to the parabola y^2- | vedclass

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

Question

(D) The given parabola is $y^2-4y-4x+8=0$. Completing the square,we get $(y-2)^2 = 4x-4$,which simplifies to $(y-2)^2 = 4(x-1)$. The equation of a tan

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(D) The given parabola is $y^2-4y-4x+8=0$. Completing the square,we get $(y-2)^2 = 4x-4$,which simplifies to $(y-2)^2 = 4(x-1)$. The equation of a tangent to the parabola $(y-k_0)^2 = 4a(x-h_0)$ with slope $m$ is $(y-k_0) = m(x-h_0) + \frac{a}{m}$. Here,$h_0=1, k_0=2, a=1$. So,$y-2 = m(x-1) + \frac{1}{m}$,which gives $y = mx - m + \frac{1}{m} + 2$. The given tangent is $x-2y+k=0$,which can be written as $y = \frac{1}{2}x + \frac{k}{2}$. Comparing the slopes,$m = \frac{1}{2}$. Comparing the intercepts,$\frac{k}{2} = -m + \frac{1}{m} + 2 = -\frac{1}{2} + 2 + 2 = \frac{7}{2}$,so $k=7$. The point on the parabola is $(1, 7)$. Differentiating the parabola $y^2-4x-4y+8=0$ with respect to $x$: $2y \frac{dy}{dx} - 4 - 4 \frac{dy}{dx} = 0$. $\frac{dy}{dx}(2y-4) = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y-2}$. At $(1, 7)$,the slope is $\frac{2}{7-2} = \frac{2}{5}$.

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the slope of the tangent drawn at $(1, k)$ on the given parabola is

Similar Questions

Let $A(1, 2)$,$B(4, -4)$,and $C(2, 2\sqrt{2})$ be points on the parabola $y^2 = 4x$. If $\alpha$ and $\beta$ respectively represent the area of $\triangle ABC$ and the area of the triangle formed by the tangents at $A, B, C$ to the parabola,then $\alpha \beta =$

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 48y$ to the ends of its latus rectum is .....

The coordinates of the focus of the parabola described parametrically by $x=5t^2+2, y=10t+4$ (where $t$ is a parameter) are

If $2x + 3y + 12 = 0$ and $x - y + 4\lambda = 0$ are conjugate with respect to the parabola $y^2 = 8x$,then $\lambda$ is equal to

The area of the triangle formed inside the parabola $y^2 = 4x$ whose vertices have ordinates $1, 2,$ and $4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module