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 \quad...\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 \quad...\left(ii\right) $
Since, parabola touches of $(x_1, y_1)$.
$ ∴$ Slope of their tangents are equal
$ − 2x_1 - 2 = - x_1 - 4$
$∴ x_1 = 2$
put the value of $x_1$ in $y_1 = -\frac{1}{2}x_1^2 - 4x_1 + 3$
we get $y_1 = -\frac{1}{2} (2)^2 - 4(2) +3 = - 7$
$ ∴ y_1 = - x_1^2 -2 x_1 + k $
$− 7 = - 4 - 4 + k $
$∴ k = 1$