Question:

The point on the parabola $y^2 = 64x$ which is nearest to the line $4x + 3y + 35 = 0$ has coordinates

Updated On: Apr 26, 2024
  • (9, -24)
  • (1, 81)
  • (4, -16)
  • (-9, -24)
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The Correct Option is A

Solution and Explanation

Given equation of parabola is
$y^{2}=64 x\,\,\,...(i)$
The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.
On differentiating both sides of E (i), we get
$2 y \frac{d y}{d x} =64$
$\Rightarrow \frac{d y}{d x} =\frac{32}{y}$
Also, slope of the given line is $-\frac{4}{3}$.
$\therefore -\frac{4}{3}=\frac{32}{y} $
$\Rightarrow y=-24$
From E (i), $(-24)^{2}=64 x$
$ \Rightarrow x=9$
$\therefore$ Hence, the required point is $(9,-24) .$
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives