The area (in sq. units) of the smaller of the | vedclass

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

Question

(B) The equation of the tangent to the parabola $y^2 = 4x$ at $(1, 2)$ is $2y = 4\left(\frac{x + 1}{2}\right)$,which simplifies to $y = x + 1$.The nor

Vedclass · Accepted Answer

$8\pi (3 - 2\sqrt{2})$

Vedclass · Answer

(B) The equation of the tangent to the parabola $y^2 = 4x$ at $(1, 2)$ is $2y = 4\left(\frac{x + 1}{2}\right)$,which simplifies to $y = x + 1$.The normal to the parabola at $(1, 2)$ has a slope of $-1$ and passes through $(1, 2)$,so its equation is $y - 2 = -1(x - 1)$,which simplifies to $y = -x + 3$.Let the center of the circle be $C(h, k)$. Since the circle touches the $x$-axis,its radius $r = |k|$. Since it also touches the normal at $(1, 2)$,the center lies on the normal,so $k = -h + 3$,or $h = 3 - k$. Thus,the center is $C(3 - r, r)$.The distance from the center $C(3 - r, r)$ to the point $P(1, 2)$ must equal the radius $r$:$PC^2 = r^2$$(3 - r - 1)^2 + (r - 2)^2 = r^2$$(2 - r)^2 + (r - 2)^2 = r^2$$2(r - 2)^2 = r^2$$2(r^2 - 4r + 4) = r^2$$r^2 - 8r + 8 = 0$Using the quadratic formula,$r = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2}$.For the smaller circle,we take $r = 4 - 2\sqrt{2}$.The area is $\pi r^2 = \pi (4 - 2\sqrt{2})^2 = \pi (16 + 8 - 16\sqrt{2}) = \pi (24 - 16\sqrt{2}) = 8\pi (3 - 2\sqrt{2})$.

The area (in $sq. units$) of the smaller of the two circles that touch the parabola $y^2 = 4x$ at the point $(1, 2)$ and the $x$-axis is

Similar Questions

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter through $A$ is

The given circle $x^2 + y^2 + 2px = 0$,$p \in R$ touches the parabola $y^2 = 4x$ externally,then

If the portion of the line $lx + my = 1$ falling inside the circle ${x^2} + {y^2} = {a^2}$ subtends an angle of $45^\circ$ at the origin,then

Tangents are drawn to the circle $x^2 + y^2 = 1$ at the points where it is met by the circles $x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3 = 0$,where $\lambda$ is a variable. The locus of the point of intersection of these tangents is:

The value of $c$ for which the set,$\{(x, y) | x^2 + y^2 + 2x \le 1 \} \cap \{(x, y) | x - y + c \ge 0\}$ contains only one point in common is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module