Question

Absolute value of slope of a line, common tangent to both the curves given by $y=x^2$ and $y=x^2+y+1=0$ will be

• A. $\sqrt{5}$
• B. $2$
• C. $\sqrt{3}$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

consider first curve $y=x^2$

Differentiate w.r.t. x, we get,

$\frac{dy}{dx}=2x$

Let common tangent touches this curve at x=a

$\therefore y=x^2=a^2$

Thus coordinates of point of contact are $(a,a^2)$

Thus, slope of tangent is $m_1=\frac{dy}{dx}=2x$

$\therefore m_1=2a$ (1)

Consider second curve $x^2+y+1=0$

Differentiate w.r.t x, we get,

$2x+\frac{dy}{dx}=0$

$\therefore \frac{dy}{dx}=-2x$

Let common tangent touches this curve at x=b

$\therefore b^2+y+1=0$

$\therefore y=-b^2-1$

Thus, coordinates of point of contact are $(b,-b^2-1)$

Thus, slope of tangent is, $m_2=\frac{dy}{dx}=-2x$

$\therefore m_2=-2b$

Now, this is common tangent to both the curves.

$\therefore m_1=m_2$

$\therefore 2a=-2b$

$\therefore a=-b$ (2)

Thus, slope of tangent is given by,

$m=\frac{y_2-y_1}{x_2-x_1}$

$\therefore m=\frac{-b^2-1-a^2}{b-a}$

From equation (2),

$\therefore m=\frac{-{(-a)}^2-1-a^2}{(-a)-a}$

$\therefore m=\frac{-a^2-1-a^2}{-2a}$

$\therefore m=\frac{-2a^2-1}{-2a}$

$\therefore m=\frac{2a^2+1}{2a}$ (3)

Equate (1) and (3), we get,

$\frac{2a^2+1}{2a}=2a$

$4a^2=2a^2+1$

$\therefore 2a^2=1$

$\therefore a^2=\frac{1}{2}$

$\therefore a=\frac{1}{\sqrt{2}}$

Thus, from equation (1),

Slope of tangent is, $m=2\times \frac{1}{\sqrt{2}}$

$\therefore m=\frac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}$

$\therefore m=\sqrt{2}$