Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If the straight-line x cosα + y sinα = p touches the curve x^2 / a^2 plus y^2/b^2 equals 1 then prove that a^2 cos2α + b^2 sin2α = p^2.

If the straight-line x \cos\alpha + y \sin\alpha = p touches the curve \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 then prove that a^2 \cos2\alpha + b^2 \sin2\alpha = p^2.

Answers (1)

Given: equation of straight-line x \cos\alpha + y \sin\alpha = p, equation of curve \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 and the straight line touches the curve

To prove: a^2 cos2\alpha + b^2 sin2\alpha = p^2.

Explanation: the know line equation is,

x \cos\alpha + y \sin\alpha = p

\Rightarrow y sin \alpha = p-x cos \alpha

\begin{align*} &\Rightarrow y = \frac{p - x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - \frac{x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - x\left(\frac{\cos \alpha}{\sin \alpha}\right) \\ &\Rightarrow y = -x\left(\frac{\cos \alpha}{\sin \alpha}\right) + \frac{p}{\sin \alpha} \\ &\text{Comparing this with the equation } y = mx + c \text{, the slope and intercept of the given line can be seen as} \\ &m = -\frac{\cos \alpha}{\sin \alpha}, \quad c = \frac{p}{\sin \alpha} \end{align*}

\begin{align*} &\Rightarrow y = \frac{p - x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - \frac{x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - x\left(\frac{\cos \alpha}{\sin \alpha}\right) \\ &\Rightarrow y = -x\left(\frac{\cos \alpha}{\sin \alpha}\right) + \frac{p}{\sin \alpha} \\ &\text{Comparing this with the equation } y = mx + c \text{, the slope and intercept of the given line can be seen as} \\ &m = -\frac{\cos \alpha}{\sin \alpha}, \quad c = \frac{p}{\sin \alpha} \end{align*}

We know that, if a line y = mx+c touches the eclipse\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, then required condition is

c\textsuperscript{2} = a\textsuperscript{2}m\textsuperscript{2}+b\textsuperscript{2}

Then putting the corresponding values, we get

\left(\frac{p}{\sin \alpha}\right)^{2} = a^{2}\left(-\frac{\cos \alpha}{\sin \alpha}\right)^{2} + b^{2} \\ \frac{p^{2}}{\sin^{2} \alpha} = \frac{\cos^{2} \alpha}{\sin^{2} \alpha}\left(a^{2}\right) + b^{2} \\ \frac{p^{2}}{\sin^{2} \alpha} = \frac{a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha}{\sin^{2} \alpha}

 
Removing the like terms we get,
\mathrm{p}^{2}=\mathrm{b}^{2} \sin ^{2} \alpha+\mathrm{a}^{2} \cos ^{2} \alpha
Hence, proved.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question