JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let the normal at a point \(P\) on the curve \(\mathrm{y}^{2}-3 \mathrm{x}^{2}+\mathrm{y}+10=0\) intersect the \(\mathrm{y}\) -axis at \(\left(0, \frac{3}{2}\right) .\) If \(\mathrm{m}\) is the slope of the tangent at \(\mathrm{P}\) to the curve, then \(|\mathrm{m}|\) is equal to
- A \(3\)
- B \(4\)
- C \(6\)
- D \(5\)
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, \(\frac{\beta-\frac{3}{2}}{\alpha}=-\frac{1}{\mathrm{m}}\)…
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