At what points on the curve y = x2 – 4x + 5 is the tangent perpendic

1.	At what points on the curve y = x2 – 4x + 5 is the tangent perpendicular to the line 2y + x = 7?
Answer» Given: The curve y = x²– 4x + 5 and line is 2y + x = 7 y = x² – 4x + 5 Differentiating the above w.r.t x, we get the Slope of the tangent, ⇒ \(\frac{dy}{dx}\) = 2x^{2 – 1}– 4 + 0 ⇒ \(\frac{dy}{dx}\) = 2x – 4 ...(1) Since, line is 2y + x = 7 ⇒ 2y = – x + 7 ⇒ y = \(\frac{-x+7}{2}\) ⇒ y = \(\frac{-x}{2}+\frac{7}{2}\) \(\therefore\) The equation of a straight line is y = mx + c, where m is the The Slope of the line. Thus, the The Slope of the line is \(\frac{-1}{2}\) ...(2) Since, tangent is perpendicular to the line, \(\therefore\) The Slope of the normal = \(\frac{-1}{\text{The Slope of the tangent}}\) From (1) & (2),we get i.e, \(\frac{-1}{2}\) = \(\frac{-1}{2x-4}\) ⇒ 1 = \(\frac{1}{x-2}\) ⇒ x – 2 = 1 ⇒ x = 3 Substituting x = 3 in y = x² – 4x + 5, ⇒ y = y = 3²– 4×3 + 5 ⇒ y = 9 – 12 + 5 ⇒ y = 2 Thus, the required point is (3,2)

At what points on the curve y = x2 – 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Answer»

Given:

The curve y = x²– 4x + 5 and line is 2y + x = 7

y = x² – 4x + 5

Differentiating the above w.r.t x,

we get the Slope of the tangent,

⇒ \(\frac{dy}{dx}\) = 2x^{2 – 1}– 4 + 0

⇒ \(\frac{dy}{dx}\) = 2x – 4 ...(1)

Since, line is 2y + x = 7

⇒ 2y = – x + 7

⇒ y = \(\frac{-x+7}{2}\)

⇒ y = \(\frac{-x}{2}+\frac{7}{2}\)

\(\therefore\) The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is \(\frac{-1}{2}\) ...(2)

Since, tangent is perpendicular to the line,

\(\therefore\) The Slope of the normal = \(\frac{-1}{\text{The Slope of the tangent}}\)

From (1) & (2),we get

i.e, \(\frac{-1}{2}\) = \(\frac{-1}{2x-4}\)

⇒ 1 = \(\frac{1}{x-2}\)

⇒ x – 2 = 1

⇒ x = 3

Substituting x = 3 in y = x² – 4x + 5,

⇒ y = y = 3²– 4×3 + 5

⇒ y = 9 – 12 + 5

⇒ y = 2

Thus, the required point is (3,2)