Question

If the parabola $y=-x^2-2x+k$ touches the parabola $y=\frac{1}{2}{x}^2-4x+3$, then the value of k is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

Let the parabola $y=-x^2-2x+k$ and the parabola $y=-\frac{1}{2}{x}^2-4x+3$ touches the point $P(x_1,y_1)$ .
Now, $y=-x^2-2x+k$
$\left(\frac{dy}{dx}\right)=-2x-2$
${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=-2x_1-2...\left(i\right)$
and $y=-\frac{1}{2}{x}^2-4x+3$
$\left(\frac{dy}{dx}\right)=-x-4$
${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=x_1-4...\left(ii\right)$
Since, parabola touches of $(x_1,y_1)$.
$\therefore$ Slope of their tangents are equal
$-2x_1-2=-x_1-4$
$\therefore x_1=2$
put the value of $x_1$ in $y_1=-\frac{1}{2}{x}_1^2-4x_1+3$
we get $y_1=-\frac{1}{2}(2{)}^2-4(2)+3=-7$
$\therefore y_1=-x_1^2-2x_1+k$
$-7=-4-4+k$
$\therefore k=1$