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For what values of \(a\) and \(b\) is the line \(2x + y = b\) tangent to the parabola \(y = a{x^2}\) when \(x = 2\)?

Short Answer

Expert verified

The values of \(a = - \frac{1}{2}\) and \(b = 2\).

Step by step solution

01

Tangent to a parabola

A tangent to a parabola is a straight line that passes through the point lying on the parabola just by touching it at that particular point without intersecting the curve.

02

Evaluating the equation of a parabola.

Thegiven equation is \(y = a{x^2}\).

Differentiating with respect to\(x\), we have:

\(y' = \;2ax\)

Now, the equation of a tangent line is \(2x + y = b\) has slope \(2\), so at \(x = 2\) we have:

\(\begin{aligned}y' &= - 2\\2a\left( 2 \right) &= - 2\\a &= - \frac{1}{2}\end{aligned}\)

Putting in the equation of parabola at\(x = 2\):

\(\begin{aligned}y &= a{x^2}\\ &= - \frac{1}{2}{\left( 2 \right)^2}\\ &= - 2\end{aligned}\)

03

Evaluating the tangent equation.

Now, for the tangent line \(2x + y = b\)at point\(\left( {2, - 2} \right)\), we have:

\(\begin{aligned}2x + y &= b\\4 - 2 &= b\\b &= 2\end{aligned}\)

So, the obtained values are:

\(\begin{aligned}a &= - \frac{1}{2}\\b &= 2\end{aligned}\)

Hence, this is the required answer.

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Most popular questions from this chapter

Find the values of a and b that make f continuous everywhere.

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^{\bf{2}}} - {\bf{4}}}}{{{\bf{x}} - {\bf{2}}}}}&{{\bf{if}}\,\,\,x < {\bf{2}}}\\{a{x^{\bf{2}}} - bx + {\bf{3}}}&{{\bf{if}}\,\,\,{\bf{2}} \le x < {\bf{3}}}\\{{\bf{2}}x - a + b}&{{\bf{if}}\,\,\,x \ge {\bf{3}}}\end{array}} \right.\)

19-24Explain why the function is discontinuous at the given number\(a\). Sketch the graph of the function.

20. \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{x + 2}}}&{if\;x \ne 2}\\1&{if\;x = - 2}\end{array}} \right.\), \({\bf{a = - 2}}\)

Explain what it means to say that

\(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{1}}^ - }} f\left( x \right) = {\bf{3}}\)and \(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{1}}^ + }} f\left( x \right) = {\bf{7}}\)

In this situation, is it possible that\(\mathop {{\bf{lim}}}\limits_{x \to {\bf{1}}} f\left( x \right)\) exists? Explain.

For the function f whose graph is given, state the value of each quantity if it exists. If it does not exist, explain why.

(a)\(\mathop {{\bf{lim}}}\limits_{x \to {\bf{1}}} f\left( x \right)\)

(b)\(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{3}}^ - }} f\left( x \right)\)

(c) \(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{3}}^ + }} f\left( x \right)\)

(d) \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{3}}} f\left( x \right)\)

(e) \(f\left( {\bf{3}} \right)\)

Each limit represents the derivative of some function f at some number a. State such as an f and a in each case.

\(\mathop {{\bf{lim}}}\limits_{x \to \frac{{\bf{1}}}{{\bf{4}}}} \frac{{\frac{{\bf{1}}}{x} - {\bf{4}}}}{{x - \frac{{\bf{1}}}{{\bf{4}}}}}\)

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