Application Of Derivatives Ques 23
23. If the curves $y^{2}=6 x, 9 x^{2}+b y^{2}=16$ intersect each other at right angles, then the value of $b$ is
(2018 Main)
(a) 6
(b) $\frac{7}{2}$
(c) 4
(d) $\frac{9}{2}$
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Answer:
Correct Answer: 23.(d)
Solution:
Formula:
Shortest distance between two curves :
- We have, $y^{2}=6 x$
$\Rightarrow \quad 2 y \frac{d y}{d x}=6 \quad \Rightarrow \quad \frac{d y}{d x}=\frac{3}{y}$
Slope of tangent at $\left(x_{1}, y_{1}\right)$ is $m_{1}=\frac{3}{y_{1}}$
Also, $\quad 9 x^{2}+b y^{2}=16$
$\Rightarrow \quad 18 x+2 b y \frac{d y}{d x}=0 \quad \Rightarrow \quad \frac{d y}{d x}=\frac{-9 x}{b y}$
Slope of tangent at $\left(x_{1}, y_{1}\right)$ is $m_{2}=\frac{-9 x_{1}}{b y_{1}}$
Since, these are intersection at right angle.
$ \begin{aligned} \therefore & m_{1} m_{2}=-1 \Rightarrow & \frac{27 x_{1}}{b y_{1}^{2}} & =1 \\ \Rightarrow & & \frac{27 x_{1}}{6 b x_{1}} & =1 \\ \Rightarrow & b =\frac{9}{2} & {\left[\because y_{1}^{2}=6 x_{1}\right] } \\ & & & \end{aligned} $
BETA
