Question

Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ is :

• A. $2x+2y-5=0$
• B. $2x+2y-3=0$
• C. $y-2=0$
• D. $none$

Answer 1

The correct option is A $2x+2y-5=0$

We have,
$y=\frac{-x^2}{2}+2$
$y^{\prime }=-x$
At $(x_1,y_1)$,
$y^{\prime }=-x_1$
The equation of a tangent at $(x_1,y_1)$ can be written as $y-y_1=-x_1(x-x_1)$
The tangent passes through $(\frac{1}{2},2)$.
Hence,
$y_1-2=-x_1(x_1-\frac{1}{2})$
$-\frac{{x_1}^2}{2}=-{x_1}^2+\frac{x_1}{2}$
$x_1=x_1^2$
$x_1=0,1$
The equation at $x_1=1,y_1=\frac{3}{2}$,
$y-\frac{3}{2}=-1(x-1)$
$2x+2y-5=0$. The distance of the line from the origin is $\frac{5}{2\sqrt{2}}<2$. Hence, this line is a secant to the circle.
The tangent at $(2,0)$ is given by $x=2$. However this is also tangent to the circle.
Hence, option A is correct.