Test Index

Parabola Part 1

Section: Mathematics

© examsnet.com

Question : 51 of 100

Marks: +1, -0

If

P

is a point on the parabola

y=x^{2}+4

which is closest to the straight line

y=4 x-1

, then the coordinates of

P

[24 Feb 2021 Shift 2]

$(3,13)$
$(1,5)$
$(-2,8)$
$(2,8)$

Solution:

Given, curve

y=x^{2}+4

and, line

y=4 x-1

Here,

y=x^{2}+4

\therefore\;\frac{dy}{dx}\;\;=2x

. . . (i)

\;\text{ and }\;y\;\;=4x-1

\;\frac{dy}{dx}\;\;=4

. . . (ii)

Let the required point be

P(x_{1}, y_{1})

.

\therefore\;\left.\frac{dy}{dx}\right|_{p}=2x_{1}

. . . (iii)

\because

Slopes will be equal.

\therefore\;\;2x_{1}=4

[from Eqs. (ii) and (iii)]

\Rightarrow

2x_{1}=4

4=2

Now, the given point

P(x_{1}, y_{1})

lies on curve

y=x^{2}+4

,

\therefore y_{1}=x_{1}^{2}+4

\Rightarrow y_{1}=2^{2}+4=8

Hence, required coordinate of

P=(2,8)

© examsnet.com

Go to Question:

Prev Question

Next Question