Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 166143 by Tawa11 last updated on 13/Feb/22

Answered by som(math1967) last updated on 14/Feb/22

 y=(2−(√x))^4    (dy/dx)=−4(2−(√x))^3 ×(1/(2(√x)))=−((2(2−(√x))^3 )/( (√x)))  [(dy/dx)]_((1,1)) =((2(2−1)^3 )/1)=−2  slope of normal through P  =(1/2)  [(dy/dx)]_((9,1)) =−((2(2−3)^3 )/3)=(2/3)  slope of normal throughQ=−(3/2)  equn. of normal throughP   y−1=(1/2)(x−1)  ⇒x−2y=−1  equn. of normal through Q  y−1=−(3/2)(x−9)  ⇒3x+2y=29  by solving x=7, y=4  R(7,4)  Area of △PQR=(1/2)∣1(1−4)+9(4−1)                                                                      +7(1−1)∣  =(1/2)×24=12sq unit

$$\:{y}=\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{4}} \\ $$$$\:\frac{{dy}}{{dx}}=−\mathrm{4}\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=−\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{3}} }{\:\sqrt{{x}}} \\ $$$$\left[\frac{{dy}}{{dx}}\right]_{\left(\mathrm{1},\mathrm{1}\right)} =\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{1}}=−\mathrm{2} \\ $$$${slope}\:{of}\:{normal}\:{through}\:{P} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left[\frac{{dy}}{{dx}}\right]_{\left(\mathrm{9},\mathrm{1}\right)} =−\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{3}\right)^{\mathrm{3}} }{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${slope}\:{of}\:{normal}\:{throughQ}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${equn}.\:{of}\:{normal}\:{throughP} \\ $$$$\:{y}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow{x}−\mathrm{2}{y}=−\mathrm{1} \\ $$$${equn}.\:{of}\:{normal}\:{through}\:{Q} \\ $$$${y}−\mathrm{1}=−\frac{\mathrm{3}}{\mathrm{2}}\left({x}−\mathrm{9}\right) \\ $$$$\Rightarrow\mathrm{3}{x}+\mathrm{2}{y}=\mathrm{29} \\ $$$${by}\:{solving}\:{x}=\mathrm{7},\:{y}=\mathrm{4} \\ $$$${R}\left(\mathrm{7},\mathrm{4}\right) \\ $$$${Area}\:{of}\:\bigtriangleup{PQR}=\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{1}\left(\mathrm{1}−\mathrm{4}\right)+\mathrm{9}\left(\mathrm{4}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{7}\left(\mathrm{1}−\mathrm{1}\right)\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{24}=\mathrm{12}{sq}\:{unit} \\ $$$$ \\ $$$$\: \\ $$

Commented by Tawa11 last updated on 14/Feb/22

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by som(math1967) last updated on 14/Feb/22

thanks for correction

$${thanks}\:{for}\:{correction} \\ $$

Commented by cortano1 last updated on 14/Feb/22

(dy/dx)= 4(2−(√x))^3 ×(−(1/(2(√x))))

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\:\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{x}}\right)^{\mathrm{3}} ×\left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com