Question

The equations of a common tangent(s) to the parabola $y={{x}^{2}}\text{ and }y=-{{\left( x-2 \right)}^{2}}$ is/ are:
This question has multiple correct options.
(a) y = 4 (x – 1)
(b) y = 0
(c) y = – 4 (x – 1)
(d) y = – 30x – 50

Answer 1

Hint : First assume the tangent of the curve in the slope intercept form. Now, substitute the equation of the line into the curve. Apply the condition of the line being tangent. Find the value of the constant in the equation of tangent in terms of the slope. Now, substitute this line in the second curve to derive the values of the slope, thus getting the constant. By this, you can get the equation of lines.

Complete step-by-step answer :
Given the equation of the curve for which we need to find the common tangent.
$y={{x}^{2}};y=-{{\left( x-2 \right)}^{2}}$

Let us assume an equation of the line with variables as m, c, we get:
y = mx + c
Let this be the common tangent to the given curves. By this statement, we can say that the line is:
y = mx + c is the tangent to the curve $y={{x}^{2}}.....\left( i \right)$
y = mx + c is the tangent to the curve $y=-{{\left( x-2 \right)}^{2}}.....\left( ii \right)$

By statement (i), we can say that if we substitute the equation of the line into the curve, we must get only one point, because tangent only touches the curve. By substituting the equation of the line into the curve, we get it as:
$mx+c={{x}^{2}}$
By subtracting mx + c on both the sides, we can write it as:
${{x}^{2}}-\left( mx+c \right)=\left( mx+c \right)-\left( mx+c \right)$
${{x}^{2}}-mx-c=0$
So, for a quadratic to have one root, the condition is to have equal roots.

& D=0 \\\ & {{b}^{2}}-4ac=0 \\\ \end{aligned}$$ Here, b = – m, a = 1, c = – c By substituting these into the condition, we get the equation: $${{m}^{2}}+4c=0$$ By simplifying we can say the value of the variable c is: $$c=\dfrac{-{{m}^{2}}}{4}....\left( iii \right)$$ By substituting this in the statement (ii), we get it as, $$y=mx-\dfrac{4}{{{m}^{2}}}$$ is the tangent to $$y=-{{\left( m-2 \right)}^{2}}$$ By substituting the equation of the line and expanding, we get it in the form of $$mx-\dfrac{{{m}^{2}}}{4}=-{{x}^{2}}-4+4x$$ By simplifying and grouping the terms, we get it in the form of $${{x}^{2}}+\left( m-4 \right)x-\left( \dfrac{{{m}^{2}}}{4}-4 \right)=0$$ This should also satisfy equal roots, D = 0, we get, $${{\left( m-4 \right)}^{2}}+4\left( \dfrac{{{m}^{2}}}{4}-4 \right)=0$$ By simplifying the whole equation, we get it in the form of: $${{m}^{2}}+16-8m+{{m}^{2}}-16=0$$ By simplifying and taking 2m common, we get it as, $$2m\left( m-4 \right)=0$$ So, the roots of m are m = 0, 4. By equation (iii), we get c to be as: $$c=-\dfrac{{{m}^{2}}}{4}=-\dfrac{{{0}^{2}}}{4},-\dfrac{{{4}^{2}}}{4}$$ So, the roots of c are c = 0, – 4. So, by these we get the equation of line to be y = 0, y = 4x – 4 or y = 4 (x – 1). Therefore, option (a) and (b) are the right answers. **Note** : Be careful while substituting the tangent into the curve as it is the only main point in the question. While applying equal roots condition, check the negative sign in a, c terms because it is being in b also doesn’t affect as we do $${{b}^{2}}.$$ While writing c in terms of m, be careful with $$-\dfrac{{{m}^{2}}}{4}$$ in confusion, students write it as $$-\dfrac{4}{{{m}^{2}}}$$ which may lead to the wrong answer.