ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 12 - 6. Application of derivatives

બિંદુ \(P\) આગળ વક્ર \(\mathrm{y}^{2}-3 \mathrm{x}^{2}+\mathrm{y}+10=0\) નો અભિલંબ \(\mathrm{y}\) -અક્ષને બિંદુ  \(\left(0, \frac{3}{2}\right) \)આગળ છેદે છે. જો \(\mathrm{m}\) એ બિંદુ \(\mathrm{P}\) આગળના વક્રના સ્પર્શકનો ઢાળ હોય તો  \(|\mathrm{m}|\) મેળવો.

  1. A \(3\)
  2. B \(4\)
  3. C \(6\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(4\)

Step-by-step Solution

Detailed explanation

Let \(\mathrm{P}(\alpha, \beta)\) so, \(\beta^{2}-3 \alpha^{2}+\beta+10=0 \ldots(i)\) Now, \(2 y y^{\prime}-6 x+y^{\prime}=0\) \(\Rightarrow \mathrm{m}=\frac{6 \alpha}{2 \beta+1}\ldots(ii)\) Also, \(\frac{\beta-\frac{3}{2}}{\alpha}=-\frac{1}{\mathrm{m}}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app