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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the points on the curve \(9y^2 = x^3\), where the normal to the curve makes equal intercepts with the axes are

\[ \begin{array}{l \qquad l \qquad l} (A)\; \left(4,\pm {\frac{8}{3}}\right) \quad &(B)\; \left(4,{\frac{-8}{3}}\right) \\ (C)\;\left(4,\pm {\frac{3}{8}}\right) \quad &(D)\;\left(\pm 4, {\frac{3}{8}}\right)\end{array} \]

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Given :
$9y^2=x^3$
Differentiating with respect to $x$
$18y.\large\frac{dy}{dx}=$$3x^2$
$\large\frac{dy}{dx}=\frac{3x^2}{18y}$
Step 2:
Let $p(x_1,y_1)$ be the point where normal is drawn.
Slope of the tangent =$\large\frac{x_1^2}{6y_1^2}$
Slope of the normal =$\large\frac{-6y_1^2}{x_1^2}$
Normal make equal intercepts on the curve
Its slope =$\pm 1$
$\large\frac{-6y_1}{x_1^2}$$=\pm 1$
$6y_1=\pm x_1^2$----(1)
Step 3:
$(x_1,y_1)$ lies on the curve $9y^2=x^3$
$9y_1^2=x_1^3$------(2)
Taking +ve sign,eliminating $y_1$ from (1) & (2)
$9.\big(\large\frac{x_1^2}{6}\big)^2=$$x_1^3$
$\large\frac{9x_1^4}{36}$$=x_1^3$
$x_1=4$
Step 4:
$y_1=\pm \large\frac{x_1^2}{6}$
by substituting the value of $x_1=4$ we get
$\quad=\pm\large\frac{16}{6}$
$\quad=\pm\large\frac{8}{3}$
The point $P$ is $(4,\pm\large\frac{8}{3})$
Part (A) is the correct answer.
answered Aug 13, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

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