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MathematicsDifficultMCQAP EAMCET · 2022
શંકુ $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ ના નાભિલંબની લંબાઈ કેટલી છે?
A
$\frac{1}{5}$
B
$\frac{2}{5}$
C
$\frac{3}{5}$
D
$\frac{4}{5}$
Solution
(D) આપેલ સમીકરણ $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ છે.
$25$ વડે ભાગતા,આપણને $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$ મળે છે.
આ $SP^2 = PM^2$ સ્વરૂપમાં છે,જ્યાં $S(2,3)$ નાભિ છે અને $3x-4y+7=0$ નિયામિકા છે.
અહીં,નાભિથી નિયામિકાનું અંતર $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}\right| = \left|\frac{6-12+7}{5}\right| = \frac{1}{5}$ છે.
પરવલયના નાભિલંબની લંબાઈ $4a$ છે.
તેથી,લંબાઈ $= 4 \times \frac{1}{5} = \frac{4}{5}$.
$25$ વડે ભાગતા,આપણને $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$ મળે છે.
આ $SP^2 = PM^2$ સ્વરૂપમાં છે,જ્યાં $S(2,3)$ નાભિ છે અને $3x-4y+7=0$ નિયામિકા છે.
અહીં,નાભિથી નિયામિકાનું અંતર $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}\right| = \left|\frac{6-12+7}{5}\right| = \frac{1}{5}$ છે.
પરવલયના નાભિલંબની લંબાઈ $4a$ છે.
તેથી,લંબાઈ $= 4 \times \frac{1}{5} = \frac{4}{5}$.
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