logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Choose the correct answer in the line \(y = mx + 1\) is a tangent to the curve \(y^2 = 4x\) if the value of \(m\) is

\[(A)\; 1 \quad (B)\; 2 \quad (C)\; 3 \quad (D)\;\frac{1}{2}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • Equation of tangent $y=mx+1$
Step 1:
The equation of the curve is $y^2=4x$
Differentiating with respect to $x$
$2y\large\frac{dy}{dx}$$=4$
$\large\frac{dy}{dx}=\frac{4}{2y}$
$\large\frac{dy}{dx}=\frac{2}{y}$
Step 2:
Slope of the tangent $=\large\frac{2}{y}$$=m$-----(1)
$(x_1,y_1)$ lies on $y^2=4x$
$y_1^2=4x_1$-----(2)
Equation of tangent $(x_1,y_1)$
$y-y_1=m(x-x_1)$-----(3)
$y=mx-mx_1+y_1$
We are given the equation of tangent
$y=mx+1$------(4)
Step 3:
Comparing (3) and (4)
$y_1-mx_1=1$------(5)
From (1) & (2)
$m=\large\frac{2}{y_1}$
$x_1=\large\frac{y_1^2}{4}$
Step 4:
Put these value in equation (5)
$y_1-\large\frac{2}{y_1}.\frac{y_1^2}{4}$$=1$
$y_1-\large\frac{y_1}{2}$$=1$
$\large\frac{2y_1-y_1}{2}$$=1$
$\large\frac{y_1}{2}$$=1$
$\Rightarrow y_1=2$
$m=\large\frac{2}{y_1}=\frac{2}{2}$$=1$
Part (A) is the correct answer.
answered Aug 13, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...